Could please answer these questions? Thanks PRACTICE IT Use the worked example a
ID: 1873473 • Letter: C
Question
Could please answer these questions? Thanks
PRACTICE IT Use the worked example above to help you solve this problem. An Eskimo returning from a successful fishing trip pulls a sled loaded with salmon. The total mass of the sled and salmon is 50.0 kg, and the Eskimo exerts a force of 1.00 x 102 N on the sled by pulling on the rope. (a) How much work does he do on the sled if the rope is horizontal to the ground (-001n the figure) and he pulls the sled 5.10 m? (b) How much work does he do on the sled if = 30.00 and he pulls the sled the same distance? (Treat the sled as a point particle, so detalls such as the polnt of attachment of the rope make no difference.) (c) At a coordinate position of 5.10 m, the Eskimo lets up on the applied force. A friction force of 46.0 N between the ice and the sled brings the sled to rest at a coordinate position of 10.10 m. How much work does friction do on the sled? EXERCISE HINTS: GETTING STARTED II'M STUCK! Suppose the Eskimo is pushing the same 50.0-kg sled across level terrain with a force of 53.5 N (a) If he does 3.00 x 10') of work on the sled while exerting the force horizontally, through what distance must he have pushed it? (b) If he exerts the same force at an angle of 30.0° with respect to the horizontal and moves the sled through the same distance, how much work does he do on the sled?Explanation / Answer
Practice it)
a) Workdone = F*s*cos(theta)
= 1*10^2*5.1*cos(0)
= 510 J
b) Workdone = F*s*cos(theta)
= 1*10^2*5.1*cos(30)
= 442 J
c) Workdone by friction, W = F*s*cos(theta)
= 46*(10.1 - 5.1)*cos(180)
= -230 J
Excercise
a) W = F*s*cos(theta)
s = W/(F*cos(theta))
= 3*10^2/(53.5*cos(0))
= 5.61 m
b) W = F*s*cos(theta)
= 53.5*5.61*cos(30)
= 260 J
Practice it)
Workdone by friction = change in kinetic energy of the car
fk*s*cos(180) = Kf - Ki
8.06*10^3*s*cos(180) = 0 - (1/2)*1.09*10^3*35^2
==> s = (1/2)*1.09*10^3*35^2/(8.06*10^3)
= 82.8 m
Excercise)
Workdone by friction = change in kinetic energy of the car
fk*s*cos(180) = Kf - Ki
8.06*10^3*32*cos(180) = 0 - (1/2)*1.09*10^3*v^2
v = 21.8 m/s