# How much work would it take to horizontally accelerate an object with a mass of 3 kg to 7 m/s on a surface with a kinetic friction coefficient of 3 ?

Dec 28, 2017

$\mu m g \Delta x + \frac{1}{2} m {v}^{2}$

#### Explanation:

Use work-energy theorem.

${W}_{\text{net}} = \Delta K$

If you ask how much 'net' work is done, then the answer is

${W}_{\text{net}} = \frac{1}{2} m {v}^{2} - \frac{1}{2} m {v}_{0}^{2} = \frac{1}{2} m {v}^{2} - 0 = \frac{1}{2} m {v}^{2}$

However, if you ask how much work is needed to overcome the friction to achieve $v = 7 \frac{m}{s}$, then you need to consider the net force and the distance traveled ($\Delta x$):

W_"net"= F_"net" Deltax = (F_"applied force" - F_mu)Deltax =DeltaK

${F}_{\mu} = \mu N = \mu M g$ is the frictional force

Therefore,

${F}_{\text{applied force}} \Delta x = {F}_{\mu} \Delta x + \Delta K = \mu m g \Delta x + \frac{1}{2} m {v}^{2}$
The workdone by the applied force therefore does two things: to overcome the frictional work and to change the kinetic energy of the object.

However, neither x nor time t to reach $v = 7 \frac{m}{s}$ are given. You need either one to solve this problem unless you meant net work done.