Question

To push a 25.0kg crate up a frictionless incline angled at25.0'to the horizontal , a worker exerts a force of 209N parallelto the incline. As the crate slides 1.5m, how much work is done onthe crate by

a) T he worker's applied force,

b) T he gravitational force on the crate, and

c) The normal force exerted by the incline on the crate?

d) What is the total work done on the crate?

Answer 1

Solution -209x1.5 work done 313.5 J b) gravitational force is acting downward. its component along the incline is, mx gxsin(25) but is acting opposite to the motion of the crate so work done by this will be negative work- -mx gx sin(25)x1.5 W155.31J the normal force exerted by the incline on the crateis, here θ1s the angle between the force and the di splacem ent. -N xd cos 90 The normal force exerted by the incline on the crateis, d) total work -Work done by the weighttwork done by the force applied +work done by normal reaction - 313.5 +0-155.31 158.18 J

Answer 2

a. work done= 209*1.5=313.5 J
b. gravitational force is acting downward. its component along the incline is m*g*sin(25)
but is acting opposite to the motion of the crate. so work done by this will be negative.
work= -m*g*sin(25)*1.5
=-155.47 J
c.the normal force exerted by the incline on the crate= m*g*cos(25)=222.27
d.total work= 313.5-155.47=158 J