Question

5.12 A 5 kg block is placed on a 10 m long incline which is at an angle of 30° to the horizontal as shown in Figure 5.9. The block is initially stationary and slides down the whole length of the incline. block o 10 m 30° Figure 5.9 If the block slides along the incline with- out friction: (a) What is the component of the grav- itational force in direction of mo- tion? (b) What is the work done by gravity on the block? (c) What is the change in gravitational potential energy? (d) What is the change in kinetic en- ergy?

How to solve from part F to L

Answer 1

Question F

The gravitational component in the direction of the motion is the sine component of the block's weight

Fm = mg sin A = 5 x 9.81 x sin 30 = 24.525N

Question G

The frictional force is the product of the block's weight into the ramp and the coefficient of friction.

F = mg cos e xu=5 x 9.81 x cos 30 X 0.462 = 19.63N

Question H

The work done by gravity is the vertical displacement times the force exerted by gravity, which is the weight of the block.

The vertical displacement is the height of the block.

h = sin 30 x 10 = 5m

Therefore the work done by gravity is

W, = mgh = 5 x 9.81 x 5 = 245.25J

Question I

The work done by the friction is the friction force times distance it has moved along the ramp, equal to the length of the slope. This work will be negative, as the friction acts opposite to the direction of the motion.

W = Fid = 19.63 x 10 = 196.3N

Question J

The change in gravitational energy will be the negative of the work done by gravity

AU,= -W, = -245.25J

Question K

The change in kinetic energy will be the sum of the work done by gravity and the work done by the friction force.

AK = W+W;= 245.25 – 196.3 = 48.95J

Question L

To find the speed, we use the relation between the kinetic energy and the speed.

AK = -mu?

48.95 = = x 5 x02

v = 4.425m/s