Question

A 65.2-kg skier coasts up a snow-covered hill that makes an angle of 33.3 ° with the horizontal. The initial speed of the skier is 8.05 m/s. After coasting a distance of 1.26 m up the slope, the speed of the skier is 4.54 m/s. (a) Find the work done by the kinetic frictional force that acts on the skis. (b) What is the magnitude of the kinetic frictional force?

Answer 1

Answer:

a) -999J

b) -793N

Explanation:

We need to apply the conversation of energy equation, we will take the potencial energy at the bottom as zero.

$u1+k1+W_{friction}=u2+k2$

now rearranging the equation:

$W_{friction}=u2+k2-k1\\W_{friction}=m*g*h+\frac{1}{2}*m*(v_2)^2-\frac{1}{2}*m*(v_1)^2\\W_{friction}=65.2*1.26*9.80*sin(33.3)+\frac{1}{2}*65.2*(4.54)^2-\frac{1}{2}*65.2*(8.05)^2\\W_{friction}=-999J$

we know that work is:

$W=F*d\\F=\frac{W}{d}\\\\F=\frac{-635}{1.26}\\F=-793N$