Jim-Bob pulls on a rope with a force of 70.1 N. The rope is attached to a sled,
ID: 1703185 • Letter: J
Question
Jim-Bob pulls on a rope with a force of 70.1 N. The rope is attached to a sled, so that the rope is 33.4 degrees above horizontal. The sled has a mass of 24.3 Kg, and the coefficient of friction between the dirt and the ground is 0.268.a. what is the acceleration of the sled?
b. how far did the sled slide in the first minute?
c. how much work did friction do on the sled in the first minute?
d. how much work did Jim-Bob do in the first minute?
e. how much work did gravity do in the first minute?
f. what was the average power of friction?
Explanation / Answer
The applied force F = 70.1N
The mass of the sled m = 24.3kg
(a) From Newton's law
ma = Fcos - [ mg - Fsin]
a ={ Fcos - [ mg - Fsin]} /m
= { (70.1)cos33.4 - (0.268)[ 24.3*9.8 - (70.1)sin33.4]} /24.3
= 0.207 m/s^2
(b) The distance traveled by sled in one min
x = (1/2) at^2
= (0.5) (0.207)(60)^2
= 372.6 m
(c) The work done by frictional force
W = f*x
= [ mg - Fsin] * x
= (0.268)[ 24.3*9.8 - (70.1)sin33.4] * (372.6)
= 19926.55 J
(d) The work done by Jim bob
W = Fcos*x
= (70.1)cos33.4 * 372.6
= 21805.6 J
(e) the work done by gravity
W = 0
(f) the avgerage power of the friction
P = W/t
= 19926.55 J / 60s
= 332 W
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