Question

The motor of a four wheeler traveling along a muddy trail generates an average power of 7.48 104 W when moving at a constant speed of 13 m/s. When pulling a log along the trail at the same speed, the engine must generate an average power of 8.50 104 W. What is the tension in the tow rope pulling the log?

Answer 1

Answer:

The tension in the tow rope pulling the log is 784.61 N

Explanation:

Given:

v = speed = 13 m/s

Pf = final power when the log is pulled = 8.5x10⁴W

Pi = average power = 7.48x10⁴W

The force required to move a car is equal to:

$F_{i} =\frac{P_{i} }{v} =\frac{7.48x10^{4} }{13} =5753.85 N$

$F_{f} =\frac{P_{f} }{v} =\frac{8.5x10^{4} }{13} =6538.46 N$

T = Ff - Fi = 6538.46 - 5753.85 = 784.61 N

Answer 2

Answer: 784.6 N

Explanation:

Given

Power of the motor, P1 = 7.48*10^4 W

Speed of the motor, v = 13 m/s

Power of the motor, P2 = 8.5*10^4 W

It is known that

Power = J/t, where

J = work, in joules

t = time, in seconds. Also,

Work = F * d, where

F = force

d = distance, thus on substituting, we have,

Power = Fd/t

Also, we know that

Velocity = d/t, where

d = distance,

t = time, thus, we could say that

Power = Force * Velocity

P = FV

To find the tension in the log, we use

F = P/V, where

P = P2 - P1

P = 8.5*10^4 - 7.48*10^4

P = 1.02*10^4 = 10200W

Force, F = 10200 / 13

F = 784.6 N