0.19 Phy1170: energy -- work  (Page 6/7)

Answer:

(10 newtons) * 5 meters = 50 joules

Note that the mass of the object is superfluous in this problem.

A force at 30 degrees

A force of 10 N is applied to a 15-kg body at an angle of 30 degrees relative to the horizontal causing the body to move 5 m to the right at constant speed. How much work is doneon the body?

Answer:

(10 N) * 5 m * cos(30 degrees) = 43.3 joules

Once again, the mass of the object is superfluous.

A vertical force

The mass is not superfluous for this scenario.

A force is applied to a 10-kg body at an angle of 90 degrees relative to the horizontal causing the body to move 5 m straight up at constant velocity. Howmuch work is done on the body?

Answer:

Weight = (10 kg) * 9.81 * (m / (s^2)) = 98.1 newtons

Under the constant velocity assumption explained earlier, an upward force equal to the weight of the object will cause theobject to continue moving in an upward direction, and eventually move 5 m straight up. The angle between the action line of the force and the displacement is 0 degrees. Therefore, the work done to theobject is:

(98.1 N) * 5 m * cos(0 degrees) = 490.5 joules

Multiple forces on a body

A block on a friction-free surface

A block on a friction-free surface is subjected to three forces:

1. Weight = 15 N
2. Upward force from the surface = 15 N
3. A 5 N force to the right that causes the block to move 5 m to the right at constant velocity.

How much work does each force do on the block.

Answer:

Forces 1 and 2 are perpendicular to the direction of displacement, so they cannot do work on the block.

Force 3 is in the same direction as the displacement and does the following work on the block:

(5 N) * 5 m * cos(0 degrees) = 25 joules

A moving block encounters friction

A block is sliding to the right on a friction free surface. Suddenly the block encounters an area on the surface where the friction is not zero. Thefriction causes the block to stop after moving 5 m to the right.

When the block encounters the frictional surface, it is subjected to the following three forces:

1. Weight = 15 N
2. Upward force from the surface = 15 N
3. A 5 N force to the left caused by the friction.

How much work does each force do on the block?

Answer:

Forces 1 and 2 are perpendicular to the direction of displacement, so they cannot do work on the block.

The angle between force 3 and the direction of displacement is 180 degrees. Force 3 does the following work on the block:

(5 N) * 5 m * cos(180 degrees) = -25 joules

Note that in this case, the work done is negative.

Another friction scenario

A block on a frictional surface is subjected to four forces:

1. Weight = 15 N
2. Upward force from the surface = 15 N
3. A 5 N force to the right causes the block to move 5 m to the right at constant speed.
4. A 5 N force to the left caused by friction.

How much work does each force do on the block?

Answer:

Forces 1 and 2 are perpendicular to the direction of displacement, so they cannot do work on the block.

The work done by force 3 is:

(5 N) * 5 m * cos(0 degrees) = 25 joules

The work done by force 4 is:

(5 N) * 5 m * cos(180 degrees) = -25 joules