A 70 kg block of ice slides down an incline 5 m long and 3 m high. A man pushes

Question

A 70 kg block of ice slides down an incline 5 m long and 3 m high. A man pushes upward on the ice so that it slides down at constant speed. The coefficient of friction between the ice and the incline is 0.1. Find a) the force exerted by the man b) the work done by the man on the block c) the work done by gravity on the block d) the work done by the surface of the incline on the block e) the work done by the resultant force on the block f) the change in kinetic energy of the block The wheels of a bus accelerating from rest acquire an angular velocity of 600 pm in 2.0 seconds. The diameter of the wheels is 1 meter. a) What is the angular acceleration of the wheels? b) Determine number of revolutions made by the wheels in this time, and the distance the bus has traveled. A uniform disk of radius .25 m and mass 5 kg is mounted on an axle supported in fixed frictionless bearings. A light cord is wrapped around the rim of the wheel and downward pull of 25N is exerted on the cord. Find the tangential acceleration of a point on the rim.

Explanation / Answer

1) Given

m = mass of the ice block = 70kg
= coefficient of friction between ice and incline = 0.1
F = force exerted by man
N = normal force on ice block
g = acceleration due to gravity (9.81 m/s2)
= incline angle where tan() = 3/5, so = 30.96°

Now if sum the forces in two directions (one parallel with the incline and one perpendicular with the incline), you will get the following 2 equations:

Parallel to incline: *N + F = m*g*sin()
Perpendicular to incline: N = m*g*cos()

a) Now combine these 2 equations to solve for F to get
F = m*g*(sin() - *cos()) = 70*9.81*(sin(30.96) - 0.1*cos(30.96))
F = 294.4 N

b) Work = force * distance = F*D
D = sqrt(32+52) = 5.83m
Since the block moves opposite the direction of the force exerted by the man, the work done is negative
W = -F * D = -242.42 * 5.83 = -1716.35 N-m

c) Since the block moves in the same direction as the force of gravity, the work done is positive
W = m*g*cos() * D = 3433.1 N-m

d) Since the block moves opposite the direction of the force exerted by friction, the work done is negative
W = *N * D = *m*g*sin() = 35.33 N-m

e) The resultant force in the direction the block is moving is zero (moving at constant velocity), so the work done by the resultant force is zero.

f) Kinetic energy is defined as (1/2)*mv2. Since the velocity is never changing as the block moves, there is zero change in kinetic energy.

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