Question

1.A box is given a push so that it slides across the floor.How far will it go, given that the coefficient of kinetic friction is 0.20 and the push imparts an initial speed of 3.0m/s?

2.A 14.0 −kg box is released on a 38 ∘ incline and accelerates down the incline at 0.25 m/s2 .

a.Find the friction force impeding its motion.

b.What is the coefficient of kinetic friction?

3.Consider the system shown in the figure.Block A has weight 5.4 N and block B has weight 3.2 N . Once block B is set into downward motion, it descends at a constant speed. Assume that the mass and friction of the pulley are negligible. Calculate the coefficient of kinetic friction μ between block A and the table top.

4.A force of 36.8 N is required to start a 4.0 −kg box moving across a horizontal concrete floor.

a.What is the coefficient of static friction between the box and the floor?

b.If the 36.8 N force continues, the box accelerates at 0.65 m/s2 . What is the coefficient of kinetic friction?

5.A car can decelerate at -4.20 m/s2 without skidding when coming to rest on a level road.

a.What would its deceleration be if the road were inclined at 11 ∘ uphill? Assume the same static friction coefficient.

Answer 1

1.

Friction force on block will be:

Ff = -uk*N = -uk*m*g

Ff = -0.20*m*g

de-acceleration of block will be, Using force balance

Ff = Fnet

-uk*m*g = m*a

a = -uk*g = -0.2*9.80 = -1.96 m/sec^2

Now Using 3rd kinematic equation:

V^2 = U^2 + 2*a*S

S = Stopping distance = (V^2 - U^2)/(2*a)

V = final speed = 0 m/sec

U = initial speed = 3.0 m/sec

So,

S = (0^2 - 3.0^2)/(2*(-1.96))

S = 2.30 m = distance traveled by box

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