Answer to Question #102548 in Mechanics | Relativity for Jarred

Question

25lb brick falls 4ft what is the force of impact?

Answer 1

Solution. According to the conditions of the problem m=11.3398kg and h=1.2192m. Using the law of conservation of mechanical energy for the initial and final positions, we write

$$mgh=\frac{mv^2}{2}$$

Hence velocity at the time of impact is

$$v=\sqrt{2gh}$$

where h is height; g=9.8 m/s^2 is gravitaty acceleration.

On the other hand, using the law of conservation of momentum, we obtain

$$m\Delta v=Ft$$

where ∆v is velocity change; F is force of impact; t is impact time. The equations of motion of a brick can be written

$$h=\frac{gt^2}{2} \implies t=\sqrt{\frac{2h}{g}}$$

Сombining the equations we get

$$F=\frac{m\Delta v}{t}=\frac{m\sqrt{2gh}}{\sqrt{\frac{2h}{g}} }=mg$$

$$F=11.3398\times 9.8\approx 111.13N$$

Answer. 111.13N