Question

A 49 kg bear slides, from rest, 11 m down a lodgepole pine tree, moving with a speed of 3.3 m/s just before hitting the ground. (a) What change occurs in the gravitational potential energy of the bear-Earth system during the slide? (b) What is the kinetic energy of the bear just before hitting the ground? (c) What is the average frictional force that acts on the sliding bear?

Answer 1

a)
`\Delta U_g=-5.3kJ`

b)
`K=0.27kJ`

c)
`F_f=0.45kN`

Step-by-step explanation:

the gravitational potential energy is given by:

`U_g=m.g.h\\`

`\Delta U_g=m.g.h_f-m.g.h_i\\\Delta U_g=49kg*9.8m/s^2*(0m-11m)\\\Delta U_g=-5.3kJ`

The kinetic energy is given by:

`K=(1)/(2)m.v^2\\`

the initial kinetic energy is zero because the motion started from rest, so:

`K=(1)/(2)*49kg*(3.3m/s^2)^2\\K=0.27kJ`

applying the conservation of energy theorem:

`U_g-W_f=K_f\\W_f=-(\Delta K+\Delta U)\\W_F=5.3kJ-0.27kJ\\W_F=-5.0kJ`

The work done by the friction force is given by:

`W_f=F_f.h.cos(\theta)\\`

the angle of the force is 180 degrees because it's against the movement:

`F_f=(W_f)/(h.cos(\theta))\\\\F_f=(-5.0kJ)/(11m.cos(180^o))\\\\F_f=0.45kN`