friction force down the plane:
mu*mg*Cos20
gravity force down the plane:
mg*sin20
work done by 100N force:
100*5=500J
work=force*distance
a. work done by gravity:
mg*sin20*(-5)=-5mg*sin20
b. work done by 100N force:
500J
c.KE change:
500J-5mg*sin20-mu*mg*cosTheta*5
d. speed final
1/2 m v^2=KEchange
solve for v
A 10 kg crate is pulled up a rough incline with an initial speed of 1.5 m/s. The pulling force is 100 N parallel to the incline, which makes an angle of 20 degrees with the horizontal. If the coefficient of kinetic friction is .40 and the crate is pulled a distance of 5 m, (a) how much work is done by gravitational force? (b) How much work is done by the 100 N force? (c) What is the change in kinetic energy of the crate? (d) What is the speed of the crate after it is pulled 5m?
1 answer
1 answer