# An object with a mass of 15 kg is moving at 9 m/s over a surface with a kinetic friction coefficient of 4 . How much power will it take to accelerate the object at #3 m/s^2?

Jan 13, 2018

Frictional force acting on the object is $f k$= $u \cdot N$ or, umg i.e 600 N,
So, let's assume we will be requiring a force of F to accelerate the object at 3 $\left(\frac{m}{\sec} ^ 2\right)$
So,using equation of force we can write,
$\left(F - f k\right)$ = $m a$
Or, F = $\left(15 \cdot 3\right)$+600 N i.e 645 N
Now,if this force cause displacement s of the object wi th in time t,power will be (work done/time) i.e 645$\left(\frac{s}{t}\right)$

Jan 13, 2018

The power is $= 5.697 k W$

#### Explanation:

The mass of the object is $m = 15 k g$

The speed is $u = 9 m {s}^{-} 1$

The acceleration is $a = 3 m {s}^{-} 2$

The coefficient of kinetic friction is

${\mu}_{k} = {F}_{r} / N = 4$

The normal force is $N = 15 g N$

The frictional force is ${F}_{r} = {\mu}_{k} \times N = 4 \cdot 15 g = 60 g N$

The force necessary to accelerate the object is $= F N$

The acceleration due to gravity is $g = 9.8 m {s}^{-} 2$

According to Newton's Second Law

$F - {F}_{r} = m a$

$F = m a + {F}_{r} = \left(\left(15 \times 3\right) + \left(60 g\right)\right) N = 633 N$

$\text{Power"="Force"xx "velocity}$

The power is

$P = F \times v = 633 \cdot 9 = 5697 W$