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

a) -999J

b) -793N

Step-by-step 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+(1)/(2)*m*(v_2)^2-(1)/(2)*m*(v_1)^2\\W_(friction)=65.2*1.26*9.80*sin(33.3)+(1)/(2)*65.2*(4.54)^2-(1)/(2)*65.2*(8.05)^2\\W_(friction)=-999J`

we know that work is:

`W=F*d\\F=(W)/(d)\\\\F=(-635)/(1.26)\\F=-793N`