A variable force of magnitude $F ( x )$ moves a body of mass $m$ along the $x$-axis from $x _ { 1 }$ to $x _ { 2 }$. The net work done by the force in moving the body from $x _ { 1 }$ to $x _ { 2 }$ is $W = \int _ { x _ { 1 } } ^ { x _ { 2 } } F ( x ) d x = \frac { 1 } { 2 } m _ { 2 } { } ^ { 2 } - \frac { 1 } { 2 } m v _ { 1 } ^ { 2 }$, where $v _ { 1 }$ and $v _ { 2 }$ are the body's velocities at $x _ { 1 }$ and $x _ { 2 }$. Knowing that the work done by the force equals the change in the body's kinetic energy, solve the problem.
-How many foot-pounds of work does it take to throw a baseball 20 mph? A baseball weighs 5 oz, or 0.3125 lb.
A) 2 ft-lb
B) 63 ft-lb
C) 4 ft-lb
D) 131 ft-lb

Question

Quizplus · Accepted Answer

To find the work done, we first convert the velocity from mph to ft/s: $20 \text{ mph} = 29.33 \text{ ft/s}$. The mass of the baseball is $0.3125 \text{ lb}$, which is equivalent to $0.3125/32.2 \text{ slugs}$. The work done is the change in kinetic energy: $W = \frac{1}{2} m v^2 = \frac{1}{2} \times \frac{0.3125}{32.2} \times (29.33)^2 \approx 4 \text{ ft-lb}$.

A Variable Force of Magnitude F(x)F ( x )F(x) Moves a Body of Mass

A Variable Force of Magnitude $F ( x )$ Moves a Body of Mass