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Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 90.0 kg, down a 60.0 ° slope at constant speed, as shown below. The coefficient of friction between the sled and the snow is 0.100. (a) How much work is done by friction as the sled moves 30.0 m along the hill? (b) How much work is done by the rope on the sled in this distance? (c) What is the work done by the gravitational force on the sled? (d) What is the total work done?

The figure is an illustration of a person in a sled on a slope that forms an angle of 60 degrees with the horizontal. Three forces acting on the sled are shown as vectors: w points vertically down, f and T point upslope, parallel to the slope.
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A constant 20-N force pushes a small ball in the direction of the force over a distance of 5.0 m. What is the work done by the force?

100 J

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A toy cart is pulled a distance of 6.0 m in a straight line across the floor. The force pulling the cart has a magnitude of 20 N and is directed at 37 ° above the horizontal. What is the work done by this force?

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A 5.0-kg box rests on a horizontal surface. The coefficient of kinetic friction between the box and surface is μ K = 0.50 . A horizontal force pulls the box at constant velocity for 10 cm. Find the work done by (a) the applied horizontal force, (b) the frictional force, and (c) the net force.

a. 2.45 J; b. – 2.45 J; c. 0

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A sled plus passenger with total mass 50 kg is pulled 20 m across the snow ( μ k = 0.20 ) at constant velocity by a force directed 25 ° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.

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Suppose that the sled plus passenger of the preceding problem is pushed 20 m across the snow at constant velocity by a force directed 30 ° below the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.

a. 2.22 kJ; b. −2.22 kJ; c. 0

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How much work does the force F ( x ) = ( −2.0 / x ) N do on a particle as it moves from x = 2.0 m to x = 5.0 m?

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How much work is done against the gravitational force on a 5.0-kg briefcase when it is carried from the ground floor to the roof of the Empire State Building, a vertical climb of 380 m?

18.6 kJ

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It takes 500 J of work to compress a spring 10 cm. What is the force constant of the spring?

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A bungee cord is essentially a very long rubber band that can stretch up to four times its unstretched length. However, its spring constant varies over its stretch [see Menz, P.G. “The Physics of Bungee Jumping.” The Physics Teacher (November 1993) 31: 483-487]. Take the length of the cord to be along the x -direction and define the stretch x as the length of the cord l minus its un-stretched length l 0 ; that is, x = l l 0 (see below). Suppose a particular bungee cord has a spring constant, for 0 x 4.88 m , of k 1 = 204 N/m and for 4.88 m x , of k 2 = 111 N/m . (Recall that the spring constant is the slope of the force F(x) versus its stretch x .) (a) What is the tension in the cord when the stretch is 16.7 m (the maximum desired for a given jump)? (b) How much work must be done against the elastic force of the bungee cord to stretch it 16.7 m?

A photograph of a person bungee jumping from a bridge above a river is accompanied by an illustration of the situation. The illustration shows the jumper at the his lowest position, and the bungee stretched by a distance l minus l sub zero.
(credit: Graeme Churchard)

a. 2.32 kN; b. 22.0 kJ

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A bungee cord exerts a nonlinear elastic force of magnitude F ( x ) = k 1 x + k 2 x 3 , where x is the distance the cord is stretched, k 1 = 204 N/m and k 2 = −0.233 N/m 3 . How much work must be done on the cord to stretch it 16.7 m?

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Engineers desire to model the magnitude of the elastic force of a bungee cord using the equation
F ( x ) = a [ x + 9 m 9 m ( 9 m x + 9 m ) 2 ] ,
where x is the stretch of the cord along its length and a is a constant. If it takes 22.0 kJ of work to stretch the cord by 16.7 m, determine the value of the constant a .

835 N

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A particle moving in the xy -plane is subject to a force
F ( x , y ) = ( 50 N · m 2 ) ( x i ^ + y j ^ ) ( x 2 + y 2 ) 3 / 2 ,
where x and y are in meters. Calculate the work done on the particle by this force, as it moves in a straight line from the point (3 m, 4 m) to the point (8 m, 6 m).

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A particle moves along a curved path y ( x ) = ( 10 m ) { 1 + cos [ ( 0.1 m −1 ) x ] } , from x = 0 to x = 10 π m, subject to a tangential force of variable magnitude F ( x ) = ( 10 N ) sin [ ( 0.1 m −1 ) x ] . How much work does the force do? ( Hint: Consult a table of integrals or use a numerical integration program.)

257 J

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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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