A typical car weights about $1200 \; N$. If the coefficient of rolling friction is $\mu_r = 0.015$. What horizontal force is needed to make the car move with constant speed of $72 \; km/hr$ on a level road? Also calculate the power developed by the engine to maintain this speed.

Given,

Weight of the car $(W) =1200 \; N$

Coefficient of friction $(\mu) = 0.015$

Velocity $(v) = 72\;km/hr = 20\;m/s$

Horizontal force $(F) = ?$

Power developed by the engine $(P) = ?$

Now we have,

$F = F_k + ma$    [Here, $F_k = \mu_r\;.\;mg$ = frictional force]

If the speed is uniform, then a = 0,

$F = \mu_r\;.\;mg = 0.015 * 1200 = 18\;N$

Therefore, horizontal force needed $(F) = 18\;N$

Also, Power developed by the engine $(P) = F * v$

$= 18 * 20 = 360\;N$

A light rope is attached to a block with mass $4 \; kg$ that rests on a frictionless, horizontal, and surface. The horizontal rope passes over a frictionless pulley and a block with mass $m$ is suspended from the other end. When the blocks are released, the tension in the rope is $10\;N$. Draw free body diagrams and calculate the acceleration of either block and the mass $m$ of the hanging block.

Let $m$ be the mass hanged at the other end of the rope.

Tension in the rope $(T) =10\;N$

Now for $4 \; kg$

$F = T = ma$

$a =$$ \frac{T}{m} = \frac{10}{4}$$ = 2.5\;m/s^2$

For mass $m$

$T = mg -ma$

or, $m(g -a) = T $    ⇒   $ m =$$ \frac{T}{(g - a)} = \frac{10}{10-2.5}$$ = 1.33 \; kg$

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