Question

Ugonna stands at the top of an incline and pushes a 100−kg crate to get it started sliding down the incline. The crate slows to a halt after traveling 1.50 m along the incline. (a) If the initial speed of the crate was 1.77 m/s and the angle of inclination is 30.0°, how much energy was dissipated by friction? (b) What is the coefficient of sliding friction?

Answer 1

To answer part (a), calculate the work done by friction using the equation work = force × distance. Calculate the change in kinetic energy of the crate using the equation change in kinetic energy = 0.5 × mass × (final velocity)^2 - 0.5 × mass × (initial velocity)^2 and subtract it from the work done by friction. To answer part (b), use the equation frictional force = coefficient of friction × normal force and calculate the normal force using the equation weight = mass × acceleration due to gravity.

Step-by-step explanation:

To answer part (a), we need to calculate the work done by friction. The work done by friction can be calculated using the equation work = force × distance. In this case, the force is the frictional force and the distance is the distance traveled by the crate. The work done by friction is equal to the change in kinetic energy of the crate, which can be calculated using the equation change in kinetic energy = 0.5 × mass × (final velocity)^2 - 0.5 × mass × (initial velocity)^2. Subtracting the change in kinetic energy from the work done by friction will give us the energy dissipated by friction.

To answer part (b), we can use the equation frictional force = coefficient of friction × normal force. The normal force is equal to the weight of the crate, which can be calculated using the equation weight = mass × acceleration due to gravity. By substituting the known values into these equations, we can find the coefficient of sliding friction.

Answer 2

Answer:(a)891.64 N

(b)0.7

Step-by-step explanation:

Mass of crate
`m=100 kg`

Crate slows down in
`s=1.5 m`

initial speed
`u=1.77 m/s`

inclination
`\theta =30^(\circ)`

From Work-Energy Principle

Work done by all the Forces is equal to change in Kinetic Energy

`W_(friction)+W_(gravity)=(1)/(2)mv_i^2-(1)/(2)mv_f^2`

`W_(gravity)=mg(0-h)=mgs\sin \theta`

`W_(gravity)=-mgs\sin \theta`

`W_(gravity)=-100* 9.8* 1.5\sin 30=-735 N`

change in kinetic energy
`=(1)/(2)* 100* 1.77^2=156.64 J`

`W_(friction)=156.64+735=891.645`

(b)Coefficient of sliding friction

`f_r\cdot s=W_(friciton)`

`891.645=f_r* 1.5`

`f_r=594.43 N`

and
`f_r=\mu mg\cos \theta`

`\mu 100* 9.8* \cos 30=594.43`

`\mu =0.7`