M*g = 15 * 9.8 = 147N. = Wt. of crate.
sin A = h/d = 2.5m/5.7m = 0.43860
A = 26o.
a. F = Fp = 147*sin A = Force parallel
with the incline.
b. Work = F*d =
d = 5.7m.
Work in Joules.
A 15 kg crate is pulled at constant speed a distance d= 5.7m up a frictionless ramp, to a
height h of 2.5 m above its starting point.
a) What force F must the cable exert on the
crate?
b) How much work is done on the crate by the
force F?
1 answer
1 answer