Question

An object with a mass of 10.0 kg is at rest at the top of a frictionless inclined plane 8 meters and an angle of inclination 30 degrees with the horizontal. The object is released from this position and it stops at a distance d from the bottom of the inclined plane along a horiztonal surface. The coefficient of kinetic friction for the horizontal surface is .400.

A. what is the knetic energy of the object at the bottom of the inclined plane?

B. what is the speed of the object at the bottom of the inclined plane?

C. at what horizontal distance from the bottom of the inclined plane will this object stop?

Please show work!

Answer 1

Given

Mass, $m = 10 \textup{kg}$

Length of the incline, $L = 8 \textup{m}$

Angle, $\theta = 30^{o}$

coefficient of kinetic friction, $\mu = 0.400$

A. The height of the incline is,

         $h = (8 \textup{m}) \textup{sin} 30^{\textup{o}}$

             $h = 4 \textup{m}$

From law of conservation of energy,

Potential energy (PE) = kinetic energy (KE)

$mgh = \frac{1}{2} m v^{2}$

          $KE = (10 \textup{kg} ) (9.80 \textup{m/s}^{2}) (4 \textup{m})$

                    $= 392 \textup{J}$

B. The speed of the object at the bottom of the incline is,

$392 \textup{J} =\frac{1}{2} m v^{2}$

$392 \textup{J} =\frac{1}{2} (10 \textup{kg}) v^{2}$

        $v = 8.85 \textup{m/s}$

C. The net force on the object is,

   

$-\mu mg = ma$

$a = - (0.400) (9.80 \textup{m/s}^{2})$

    $= - 3.92 \textup{m/s}^{2}$

The distance traveled by the object is,

$0- (8.85 \textup{m/s})^{2}= 2(- 3.92 \textup{m/s}^{2})d$

$d = 9.99 \textup{m}$

   $= 10.0 \textup{m}$

Answer 2

Vertical height = 8 sin30

from energy conservation:

KE = mgh = 10*9.8*8 sin30 = 392J

b) .5 mv^2 = 392

v = 8.85 m/sec

c) v^2 = 2a*d

a = .4*g

so, d = 9.99 m