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Two blocks connected by a string are pulled across a horizontal surface by a force applied to one of the blocks, as shown below. The coefficient of kinetic friction between the blocks and the surface is 0.25. If each block has an acceleration of 2.0 m/s 2 to the right, what is the magnitude F of the applied force?

Two blocks, 1.0 kilograms on the left and 3.0 kilograms on the right, are connected by a string and are on a horizontal surface. Force F acts on the 3.0 kilogram mass and points up and to the right at a angle of 60 degrees above the horizontal.

25 N

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As shown below, the coefficient of kinetic friction between the surface and the larger block is 0.20, and the coefficient of kinetic friction between the surface and the smaller block is 0.30. If F = 10 N and M = 1.0 kg , what is the tension in the connecting string?

Two blocks, 2 M on the left and M on the right, are connected by a string and are on a horizontal surface. Force F acts on M and points to the right.
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In the figure, the coefficient of kinetic friction between the surface and the blocks is μ k . If M = 1.0 kg, find an expression for the magnitude of the acceleration of either block (in terms of F , μ k , and g ).

Two blocks, M on the left and 3 M on the right, are connected by a string and are on a horizontal surface. The following forces are indicated: f sub k 2 acting on M and pointing to the right, f sub k 1 acting on 3 M and pointing to the right, F acting on 3 M and pointing to the left, N sub 2 acting on M and pointing up, N sub 1 acting on 3 M and pointing up, M g acting on M and pointing down, , 3 M g acting on 3 M and pointing down.

a = F 4 μ k g

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Two blocks are stacked as shown below, and rest on a frictionless surface. There is friction between the two blocks (coefficient of friction μ ). An external force is applied to the top block at an angle θ with the horizontal. What is the maximum force F that can be applied for the two blocks to move together?

Rectangular block M sub 2 is on a horizontal surface. Rectangular block M sub 1 is on top of block M sub 2. A force F pushes on block M sub 1. Force F is directed down and to the right, at a angle theta to the horizontal.
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A box rests on the (horizontal) back of a truck. The coefficient of static friction between the box and the surface on which it rests is 0.24. What maximum distance can the truck travel (starting from rest and moving horizontally with constant acceleration) in 3.0 s without having the box slide?

14 m

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A double-incline plane is shown below. The coefficient of friction on the left surface is 0.30, and on the right surface 0.16. Calculate the acceleration of the system.

Two carts connected by a string passing over a pulley are on either side of a double inclined plane. The string passes over a pulley attached to the top of the double incline. On the left, the incline makes an angle of 37 degrees with the horizontal and the cart on that side has mass 10 kilograms. On the right, the incline makes an angle of 53 degrees with the horizontal and the cart on that side has mass 15 kilograms.
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Challenge problems

In a later chapter, you will find that the weight of a particle varies with altitude such that w = m g r 0 2 r 2 where r 0 is the radius of Earth and r is the distance from Earth’s center. If the particle is fired vertically with velocity v 0 from Earth’s surface, determine its velocity as a function of position r . ( Hint: use a d r = v d v , the rearrangement mentioned in the text.)

v = v 0 2 2 g r 0 ( 1 r 0 r )

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A large centrifuge, like the one shown below, is used to expose aspiring astronauts to accelerations similar to those experienced in rocket launches and atmospheric reentries. (a) At what angular velocity is the centripetal acceleration 10 g if the rider is 15.0 m from the center of rotation? (b) The rider’s cage hangs on a pivot at the end of the arm, allowing it to swing outward during rotation as shown in the bottom accompanying figure. At what angle θ below the horizontal will the cage hang when the centripetal acceleration is 10 g ? ( Hint: The arm supplies centripetal force and supports the weight of the cage. Draw a free-body diagram of the forces to see what the angle θ should be.)

(a) A photograph of a high g training centrifuge. The astronaut sits in a cage at the end of a long arm that rotates in a horizontal plane. (b) An illustration of a top view of the centrifuge along with an illustration of the forces. The free body diagram shows the weight, w, pointing vertically down and the force F sub arm pointing up and to the left. The forces are then shown rearranged to form a right triangle. F sub arm is the hypotenuse of the triangle pointing up and left, w is the vertical side pointing down, and F sub c is the base pointing to the left. The F sub c arrow is then shown separately with the notation that vector F sub c equals F sub net.
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A car of mass 1000.0 kg is traveling along a level road at 100.0 km/h when its brakes are applied. Calculate the stopping distance if the coefficient of kinetic friction of the tires is 0.500. Neglect air resistance. ( Hint: since the distance traveled is of interest rather than the time, x is the desired independent variable and not t . Use the Chain Rule to change the variable: d v d t = d v d x d x d t = v d v d x . )

78.7 m

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An airplane flying at 200.0 m/s makes a turn that takes 4.0 min. What bank angle is required? What is the percentage increase in the perceived weight of the passengers?

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A skydiver is at an altitude of 1520 m. After 10.0 seconds of free fall, he opens his parachute and finds that the air resistance, F D , is given by the formula F D = b v , where b is a constant and v is the velocity. If b = 0.750 , and the mass of the skydiver is 82.0 kg, first set up differential equations for the velocity and the position, and then find: (a) the speed of the skydiver when the parachute opens, (b) the distance fallen before the parachute opens, (c) the terminal velocity after the parachute opens (find the limiting velocity), and (d) the time the skydiver is in the air after the parachute opens.

a. 53.9 m/s; b. 328 m; c. 4.58 m/s; d. 257 s

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In a television commercial, a small, spherical bead of mass 4.00 g is released from rest at t = 0 in a bottle of liquid shampoo. The terminal speed is observed to be 2.00 cm/s. Find (a) the value of the constant b in the equation v = m g b ( 1 e b t / m ) , and (b) the value of the resistive force when the bead reaches terminal speed.

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A boater and motor boat are at rest on a lake. Together, they have mass 200.0 kg. If the thrust of the motor is a constant force of 40.0 N in the direction of motion, and if the resistive force of the water is numerically equivalent to 2 times the speed v of the boat, set up and solve the differential equation to find: (a) the velocity of the boat at time t ; (b) the limiting velocity (the velocity after a long time has passed).

a. v = 20.0 ( 1 e −0.01 t ) ; b. v limiting = 20 m/s

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Questions & Answers

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Derived the formula of Newton's law of universal gravitation Fg=G(M1M2)/R2
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Of?
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centripetal force of attraction that pulls a body that is traversing round the orbit of a circle toward the center of the circle. Fc = MV²/r
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I do believe the formula for centripetal force is F=MA or F=m(v^2/r)
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I mean the formula is Fc= Mass multiplied by square of velocity all over the Radius of the circle
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The force is equal to the mass times the velocity squared divided by the radius
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That's the current chapter I'm on in my engineering physics class
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Practice Key Terms 2

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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