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Nigzzzz
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A snow sled, with total mass 80.0 kg, is lowered at constant speed down a slope of angle 60.0° with respect to the horizontal, for a distance d = 14.0 m. The coefficient of kinetic friction between the sled and the snow is 0.290.
Note: g = 9.80 m s–2.
1. What is the work done by the friction force when the sled slides a distance d down the inclined slope?
Equations tried using:
Fnet = F1 + F2 + F3
Ffr = µf(mgcos theta)
Fnet = Wsin theta - F friction
i found the Fw to be 541N and the Ff to be 124N but then i seem to get lost on how to get tension force to get the net work done by friction, i used the formula Fnet = Fw-T-Ff = 0
Fw - Ff = T = 416N then got stuck
Note: g = 9.80 m s–2.
1. What is the work done by the friction force when the sled slides a distance d down the inclined slope?
Equations tried using:
Fnet = F1 + F2 + F3
Ffr = µf(mgcos theta)
Fnet = Wsin theta - F friction
i found the Fw to be 541N and the Ff to be 124N but then i seem to get lost on how to get tension force to get the net work done by friction, i used the formula Fnet = Fw-T-Ff = 0
Fw - Ff = T = 416N then got stuck