Question

a body of mass 10kg is being acted upon by a force 3t² and an opposing constant force of 32N.The initial speed is 10m/s . Calculate the velocity of body after 5seconds.

Answer 1

Answer:

The velocity of body after 5 seconds is 6.5 m/s

Explanation:

Mass of the body m = 10 kg

Net force on the body F = 32 - 3t²

Acceleration on the body a = F/m

                                             = $\frac{3t^2-32}{10}$

At t = 0 initial velocity v₀ = 10 m/s

We have to find velocity after t = 5 sec

We know that rate of change of velocity = acceleration

$\implies \frac{dv}{dt} =a$

$\implies \frac{dv}{dt} =(3t^2-32)/10$

$\implies dv =\frac{3t^2-32}{10}dt$

or, $\int_{v_0}^v dv=(1/10) \int_0^t(3t^2-32)dt$

$\implies \int_{10}^v dv=(1/10) \int_0^5(3t^2-32)dt$

$\implies v\Bigr |_{10}^v=(1/10) (t^3-32t) \Bigr |_0^5$

$\implies v-10=\frac{1}{10} \times (5^3-32\times 5)$

$\implies v-10=\frac{1}{10} \times 5 \times (5^2-32)$

$\implies v-10=\frac{1}{2} \times (25-32)$

$\implies v=10+\frac{1}{2} (-7)$

$\implies v=10-3.5$

$\implies v=6.5 \text{ m/s}$

Therefore, the velocity of body after 5 seconds is 6.5 m/s