Wb = M*g = 200 N., M = 200/g = 200/9.8 =
20.4 kg.
sin A = 3/10 = 0.30, A = 17.5o.
Fp = 200*sin17.5 = 60 N.
Fn = 200*Cos17.5 = 191 N.
a. Work = F*d = 120 * 10 = 1200 J.
b. Mg*h-0 = Mg*h = 200 * 3 = 600 J.
c. Conservation Energy : Change in KE =
Change in PE = 600 J.
d. Fap-Fp-Fk = M*a.
120-60-Fk = M*0 = 0.
Fk = 120-60 = 60 N.
A 200 N box is pushed up an incline 10 m long and 3 m high. The average force (parallel to the plane) is 120 N.
a.) How much work is done?
b.) What is the change in the PE of the box?
c.) What is the change in KE of the box?
d.) What is the frictional force on the box?
2 answers
Roger pushes a box on a 30° incline. If he applies a force of 60 newtons parallel to the incline and displaces the box 10 meters along the incline, how much work will he do on the box?
1 answer