Question

A rocket sled accelerates at a rate of 49.0 m/s2 . Its passenger has a mass of 75.0 kg. (a) Calculate the horizontal component of the force the seat exerts against his body. Compare this with his weight using a ratio. (b) Calculate the direction and magnitude of the total force the seat exerts against his body.

Answer 1

Explanation:

It is given that,

Mass of the passenger, m = 75 kg

Acceleration of the rocket, $a=49\ m/s^2$

(a) The horizontal component of the force the seat exerts against his body is given by using Newton's second law of motion as :

F = m a

$F=75\ kg\times 49\ m/s^2$

F = 3675 N

Ratio, $R=\dfrac{F}{W}$

$R=\dfrac{3675}{75\times 9.8}=5$

So, the ratio between the horizontal force and the weight is 5 : 1.

(b) The magnitude of total force the seat exerts against his body is F' i.e.

$F'=\sqrt{F^2+W^2}$

$F'=\sqrt{(3675)^2+(75\times 9.8)^2}$

F' = 3747.7 N

The direction of force is calculated as :

$\theta=tan^{-1}(\dfrac{W}{F})$

$\theta=tan^{-1}(\dfrac{1}{5})$

$\theta=11.3^{\circ}$

Hence, this is the required solution.