Mechanics 2 - friction

用戶25071

2010-03-01 7:57 am

有冇人可以教我呢條問題呀,,!? THX!

A box of mass 20kg is being pushed along a uniform rough horizontal floor by means of a downwared force of 40N at 70 degree to the vertical. The box is initially at rest and is travelling at 0.8 ms-1 after it has slid 6m. You may assume that the box does not tip up.

i.) find the value of the frictional force, assuming that it is constant. Give your answers correct to 2 significant figures.

ii.) Calcaulate the value of the coefficient of friction

When the box is trvaelling at 0.8 ms-1, the applied force is removed.
iii.) How far does the box slide before coming to rest?

iv.) If the force applied to the box had been 40N upwards at 70 degree to the vertical and the box did not tip up, would it have been travelling at the same speed, ot faster or slower, after sliding 6m? You should give a reason for your answer but are not required o calculate the speed.

回答 (1)

Prof. Physics

2010-03-01 1:27 pm

✔ 最佳答案

i. By equation of motion, s = (u + v)t/2

6 = (0 + 0.8)t/2, Time, t = 15 s

By v = u + at

0.8 = 0 + a(15)

Acceleration, a = 0.0533 ms^-2

By Newton's 2nd law of motion,

F = ma

40sin70* - f = 20(0.0533)

Frictional force, f = 36.5 N

ii. Normal reaction, R = mg + 40cos70* = 214 N

By u = f/R

Coefficient of friction, f = 36.5 / 214 = 0.17

iii. By f = ma

-36.5 = 20a

Acceleration, a = -1.83 ms^-2

By equation of motion,

v^2 = u^2 + 2as

0 = (0.8)^2 + 2(-1.83)s

Distance, s = 0.175 m

iv. The travelling speed should be larger. The parallel force resolved from the applied force is the same, but the normal reaction on the box is smaller, and hence the friction acting on it is smaller, resulting a larger resultant force in the parallel direction.

參考: Physics king

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