Question

As a van travels along a level stretch of road with constant speed, most of the power developed by the engine is used to compensate for the energy transformations due to friction forces exerted on the van by the air and the road. If the power developed by the engine is 2.75 hp, calculate the total friction force acting on the van (in N) when it is moving at a speed of 32 m/s. One horsepower equals 746 W.

Answer 1

Answer:

The friction force on the van is 64.11 N

Solution:

As per the question:

Power developed by the van engine, P = 2.75 hp

Speed of the van, v = 32 m/s

Since, 1 hp = 746 W

Thus

P = $2.75\times 746 = 2051.5 W$

Now, we know that:

Work done is given as the product of force and displacement:

W = Fd                   (1)

where

W = Work done

F = Force

d = displacement

Also,

$P = \frac{dW}{dt}$          (2)

where

t = time

Therefore, from eqn (1) and (2):

$P = \frac{d}{t}(Fd)$

Since, the van is travelling with constant speed, Force is constant:

P = Fv

$F = \frac{P}{v} = \frac{2051.5}{32} = 64.11\ N$

• Since, this force is applied by the engine in order to maintain the speed and hence is equal to the friction force that acts on the van.

Answer 2

Answer:

64.10937 N

Explanation:

s = Displacement

v = Velocity

F = Force

Converting hp to W

$2.75\times 746=2051.5\ W$

Work

$W=F\times s$

Power

$P=\frac{W}{t}\\\Rightarrow P=\frac{F\times s}{t}$

Now

$\frac{s}{t}=v$

So,

$P=F\times v\\\Rightarrow F=\frac{P}{v}\\\Rightarrow F=\frac{2051.5}{32}\\\Rightarrow F=64.10937\ N$

The friction force acting on the van is 64.10937 N