Question

A 3kg crab was moving at 1 m/s in the shore before the ride pushed him for 5 seconds. If his final speed was 3 m/s, what force did the tide push him with?

Answer 1

`1.2\; {\rm N}`, assuming that all other forces on this crab were balanced.

Step-by-step explanation:

The impulse
`J` on an object is equal to the change in momentum
`\Delta p`. In other words:

`J = \Delta p`.

If the mass
`m` of the object stays the same (as in the case of this question), the change in momentum can be rewritten as:

`J = \Delta p = m\, \Delta v`, where
`\Delta v` is the change in velocity.

Impulse is also equal to the net force on the object
`F_{\text{net}}` times the duration
`\Delta t` over which the force is applied:

`J = F_{\text{net}}\, \Delta t`.

Equate the two expressions for
`J` to obtain:

`F_{\text{net}}\, \Delta t = m\, \Delta v`.

In this question:

• 
  `\Delta t = 5\; {\rm s}` is the duration over which the force was applied,
• 
  `m = 3\; {\rm kg}` is the mass of the crab, and
• 
  `\Delta v = (3 - 1)\; {\rm m\cdot s^(-1)} = 2\; {\rm m\cdot s^(-1)}` is the change in the velocity of the crab.

Rearrange
`F_{\text{net}}\, \Delta t = m\, \Delta v` and solve for the net force
`F_{\text{net}}`:

`\begin{aligned}F_{\text{net}} &= (m\, \Delta v)/(\Delta t) \\ &= \frac{(3\; {\rm kg})\, (2\; {\rm m\cdot s^(-1)})}{5\; {\rm s}} \\ &= 1.2\; {\rm kg \cdot m\cdot s^(-2)} \\ &= 1.2\; {\rm N}\end{aligned}`.

Assuming that all other forces on this crab are balanced, the net force on the crab would be equal to the force from the tide. Hence, the tide would have pushed the crab with a force of
`1.2\; {\rm N}`.