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a) A 2000 kg car is travelling 50 miles per hour. Find the
kinetic energy in Joules. b) The same car is lifted vertically upward and then
dropped from rest. Find the height from which it is dropped if it strikes the
ground at 50 miles per hour (neglect air resistance).

**Solution: **

**a)**-

*KE*= *mv*^{ 2}= (2 `x`10^{3}*kg*)^{2}= 4.99 `x`10^{5}*J*(10) **b)**-

*PE*_{i}= *KE*_{f}*mgh*= *mv*^{ 2}*h*= = ^{2}= 25.5*m*(11)

An object of mass 1 kg travelling at 5.0 m/s enters a region of
ice where the coefficient of kinetic friction is .10. Use the work energy
theorem to find the distance the object travels before coming to rest.

**Solution: **

The work energy theorem gives
*W* = *KE* . We have
*W* = - *f*_{k}*d* = - *Nd* = - *mgd* and
*KE* = *mv*_{f}^{2} - *mv*_{i}^{2} = - *mv*_{i}^{2}. Combining,

- mgd
| = |
- mv_{i}^{2}
| |

d
| = |
v_{i}^{2}
| |

= |
= 13 m.
| (12) |

A 30 kg child enters the final section of a waterslide travelling
at 2.0 m/s. The final section is 5.0 m long and has a vertical drop of 3.0 m.
The force of friction opposing the child's motion is 50 N. Find a) the loss of
potential energy, b) the work done by friction in the final section and c) the
child's velocity at the end of the section (using energy considerations).

**Solution: **

**a)**-

*PE*= *mg*(*h*_{f}-*h*_{i})= 30(9.8)(0 - 3) = - 882 *J*(13) **b)**-

*W*= -*f*_{k}*x*= - 50(5) = - 250*J*(14) **c)**-

*W*_{nc}= *KE*+*PE*- 250 = (30)( *v*_{f}^{2}) - (30)(2.0)^{2}- 882*v*_{f}^{2}= *v*_{f}= 6.8 *m*/*s*(15)

A 2.0 kg wood block is on a level board and held against a spring
of spring constant k=100 N/m which has been compressed .1 m. The block is
released and pushed horizontally across the board. The coefficient of friction
between the block and the board is = .20. Find a) the velocity of the block just
as it leaves the spring and b) the distance the block travels after it leaves
the spring.

**Solution: **

**a)**- The work energy theorem gives:

*W*_{nc}= *KE*+*PE*- *f*_{k}*x*= ( *mv*_{f}^{2}- 0) + (0 -*kx*^{ 2})- *mgx*= *mv*_{f}^{2}-*kx*^{ 2}*v*_{f}^{2}= = *v*_{f}= 0.33 *m*/*s*.(16) **b)**- The work energy theorem gives,

- *mgd*= 0 - *mv*_{i}^{2}*d*= *v*_{i}^{2}= = 0.028*m*.(17)

A man pushes a 100 kg box across a level floor at a constant
speed of 2.0 m/s for 10 s. If the coefficient of friction between the box and
the floor is
= 0.20 , find the average power output by the man.

**Solution: **

Since the acceleration of the box is zero the force exerted by the man is
obtained from
*F*_{ man} - *f*_{k} = 0 *F*_{ man} = *f*_{k} = *mg* .
Then

P = = F = mg = .2(100)(9.8)(2) = 392 W.
| (18) |

10/9/1997