Question

A 6.00 kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied to the end of the rope, and the height of the crate above its initial position is given by y(t)=(2.80m/s)t +(0.61 m/s3 )t3
What is the magnitude of the force F when 4.40 s ?

Answer 1

Answer:

$F=52.1301\ N$

Explanation:

Given:

• mass attached to the rope, $m=6\ kg$

• given time, $t=4.4\ s$

function of height of the mass connected via a rope:

• $y=2.8\ t+0.61\ t^3$

differentiate the above eq. with respect to time t gives us the velocity in vertical direction:

$v_y=\frac{d}{dt} y$

$v_y=2.8+1.83\ t^2$

put the value of given time:

$v_y=2.8+1.83\times 4.4^2$

$v_y=38.2288\ m.s^{-1}$

• As we know form the Newton'a second law of motion that the rate of change in momentum is proportional to the applied force.

$F=\frac{\Delta (m.v_y)}{t}$

$F=\frac{6\times 38.2288}{4.4}$

$F=52.1301\ N$