A 75.0-kg crate has been given initial speed of 4.5 m/s when it begins to slide

Question

A 75.0-kg crate has been given initial speed of 4.5 m/s when it begins to slide on a rough surface with coefficient of friction of 0.15.

A) How much mechanical energy is lost due to friction acting on the crate until the moment when its speed dropped to 2.0 m/s?

B) How far is the crate from the sliding starting position at the moment when its speed is 2.0 m/s?

C) Compared to the position when it moves at 2.0 m/s, how much further can the crate move?

D) On some other surface the same crate under the same initial conditions moves 10.0 m before it comes to a full stop. What is the coefficient of friction of this surface? How does it compare with the coefficient of friction on the original surface?

Explanation / Answer

Given: m = 75 kg, Vi = 4.5 m/s, = 0.15, Vf = 2 m/s

A) Change in kinetic energy is given by,

K.E. = ½ m(Vf² - Vi²)

          = ½ * 75 * (2² - 4.5²)

         = 0.5 * 75 * (4 – 20.25)

         = - 609.375 J

Since the energy cannot be negative as it is a scalar quantity. The answer is 609.375 J.

B) The change in velocity i.e. change in K.E. is due to force of friction. Work done is change in K.E. due to friction in this case. Distance covered during this change in K.E. is d.

So, using, W = Fs * d

i.e. 609.375 = Fs*d

So, 609.375 = * N * d

Therefore, d = 609.375/( * N)     [N = mg]

i.e. d = 609.375/(0.15*75*9.8)

d = 5.527 m (Answer)

C) The crate has a speed of 2 m/s, the maximum distance it moves when it’s speed Vf = 0

Using the same formula, from part B and Vf = 0

We get,

½* 75 * 2² = Fs * d

i.e. d = 150/110.25

d = 1.36 m

So, the crate will cover the distance of 1.36 m when it has a speed of 2 m/s.

D) Now, when the crate covers a distance of d = 10 m, before coming to rest i.e. Vf = 0 m/s

So, as per initial conditions, Vi = 4.5 m/s.

We can write,

½* 75 * 4.5² = * N * d

759.375 = * 75 * 9.8 * 10

So, = 759.375/(75*98) = 0.10 (Answer)

So when the crate covers a distance of 10 m, coefficient of friction = 0.10.

Since, compared to initial case, in part D) coefficient of friction is less, crate covers a more distance. As the frictional force reduces due to low coefficient of friction, opposing force is less, the crate covers greater distance.

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