Question

Question 2

Points 2
A 7.5 kg bowling ball, initially at rest, is dropped from the top of an 11 m building. It
hits the ground 1.5 s later. Find the net external force on the falling ball.
110 N
R
CO

Answer 1

Approximately
`73\; {\rm N}`, assuming that the acceleration of this ball is constant during the descent.

Step-by-step explanation:

Assume that the acceleration of this ball,
`a`, is constant during the entire descent.

Let
`x` denote the displacement of this ball and let
`t` denote the duration of the descent. The SUVAT equation
`x = (1/2)\, a\, t^(2)` would apply.

Rearrange this equation to find an expression for the acceleration,
`a`, of this ball:

`\begin{aligned} a &= (2\, x)/(t^(2))\end{aligned}`.

Note that
`x = 11\; {\rm m}` and
`t = 1.5\; {\rm s}` in this question. Thus:

`\begin{aligned} a &= (2\, x)/(t^(2)) \\ &= \frac{2 * 11\; {\rm m}}{(1.5\; {\rm s})^(2)} \\ &\approx 9.78\; {\rm m \cdot s^(-2)}\end{aligned}`.

Let
`m` denote the mass of this ball. By Newton's Second Law of Motion, if the acceleration of this ball is
`a`, the net external force on this ball would be
`m\, a`.

Since
`m = 7.5\; {\rm kg}` and
`a \approx 9.78\; {\rm m\cdot s^(-2)}`, the net external force on this ball would be:

`\begin{aligned} (\text{net force}) &= m\, a \\ &\approx 7.5\; {\rm kg} * 9.78\; {\rm m\cdot s^(-2)} \\ &\approx 73\; {\rm kg \cdot m \cdot s^(-2) \\ &= 73\; {\rm N} && (1\; {\rm N} = 1\; {\rm kg \cdot m\cdot s^(-2)}) \end{aligned}`.