The constant force resisting the motion of a car of mass $1500 \; kg$ is equal to one fifteenth of its weight. When travelling at $48 \; km/h$, the car is brought to rest in a distance of $50 \; m$ by applying the brakes, find the additional retarding force due to the brakes (assumed constant) and heat developed in the brakes.

Given,

Mass of car (m) = 1500 kg

Frictional force ($F_c)$ = $\frac{1}{15}$$\;m.g$

                                      = $\frac{1}{15}$$\;*\;1500\;*\;10 = 1000\;N$

Initial velocity (u) = 48 km/hr = $\frac{48\;*\;1000}{60\;*\;60}$ $= 13.33\;m/s$

Distance covered (s) = 50 m

Final velocity (v) = 0

Additional force ($F_a = \;?$)

Heat developed in the engine = ?

To calculate, $a = ?$ We have,

$v^2 = u^2 \;+\;2.a.s$

$0^2 = (13.33)^2 + 2\;*\;a\;*\;50$

⇒ a = $-$ $\frac{(13.33)^2}{2\;*\;50}$$ = -\;1.78\;m/s$

The retarding force due to friction is,

$F_r = m\;a = -\;1.78\;*\;1500 = 2666.67\;N$

But frictional force ($F_c = 1000\;N$)

So, Additional force ($F_a$) = $F_r - F_c = 2666.67 - 1000 = 1666.67\;N$

Now heat developed = ?

Heat developed = Work done

i.e. $W = F_a\;*\;s = 1666.67\;*\;50 = 83333.33\;J$

∴ Additional retarding force due to brake is $1666.67\;N$

∴ Heat developed in the engine is $83333.33\;J$

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Suppose you try to move a carate by tying a rope qround it and pulling on the rope at angle of $30^0$ above the horizontal. What is the tension required to keep the crate moving with constant velocity? Assume weight of the crate $W = 500N$ and coefficient of dynamic fraction $\mu_k = 0.40$.

Given,

Let $W$ be the weight of the crate, and $R$ be the normal reaction, $T$ be the tension and $f_s$ be the frictional force. 

Let $\theta = 30^0$ be the angle between $T$ and horizontal. Then,

W = R

In horizontal direction, $F = T\;cos\theta \; - \; F_s$

For constant velocity, $a = 0$

$T\cos\theta = F_s$

$T$ = $\frac{\mu_k\;*\;R}{cos\theta}$ = $\frac{0.4 \;*\; 500}{cos(30^0)}$ = $230.94 \; N$

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A block is pushed 1.5 m along a horizontal tabletop with horizontal force of 2.4 N. If the frictional force between the surface in contact is 0.6 N, what is (i) work done on the block by the frictional force and (ii) the total work done on the block?

Given,

Distance (d) = 1.5 N

Horizontal force ($F_H$) = 2.4 N

Frictional Force ($F_C$) = 0.6 N

(i) Work done by the frictional force ($W_f$) = $F_c * d$ = $-0.6 * 1.5$ = $-0.9 \; Joule$

(ii) Work done by the horizontal force ($W_H$) = $F_H * d$ = $2.4 * 1.5$ = $3.6 \; Joule$

Now, 

Total work done ($W$) = $W_f$ + $W_H$ = $3.6 - 0.9$ = $2.7 \; Joule$

  

A $650\;KW$ power engine of a vehicle of mass $1.5 \; * \; 10^5\; kg$ is rising on an inclined plane of inclination $1$ in $100$ with a constant speed of $60\;km/hr$. Find the frictional force between the wheels of the vehicle and the plane.

Given,

Power of engine $(P) = 650\;kw = 650000\;W$

Mass of vehicle $m = 1.5 \; * \; 10^5\; kg$

Inclination $Sin\;\theta = \frac{1}{100}$

Speed of vehicle $(v) = 60\;km/hr = $$\frac{60\;*\;1000}{60\;*\;60} = \frac{50}{3}m/s$

Frictional force $(f_r) = \;?$

We have, total upwarding force $F = f_r + mg\;sin\;\theta$    ⇒    $\frac{P}{v}$$ = f_r + mg\;sin\;\theta$

or, $f_r = \frac{P}{v} - mg\;sin\;\theta$  =  $\frac{650000\;*\;3}{50}$$ - 1.5\;*\;10^5\;*\;10\;*\;$$\frac{1}{100}$ = $24000\;N$

Hence, fractional force between the wheels of the vehicles and the plane is $24000\;N$.

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