Question

A particle of mass m=5.00 kilograms is at rest at t=0.00 seconds. a varying force f(t)=6.00t2−4.00t+3.00 is acting on the particle between t=0.00 seconds and t=5.00 seconds. find the speed v of the particle at t=5.00 seconds. express your answer in meters per second to three significant figures.

Answer 1

Answer:

The speed v of the particle at t=5.00 seconds = 43 m/s

Explanation:

Given :

mass m = 5.00 kg

force f(t) = 6.00t2−4.00t+3.00 N

time t between t=0.00 seconds and t=5.00 seconds

From mathematical expression of Newton's second law;

Force = mass (m) x acceleration (a)

F = ma              

$a = \frac{F}{m}$      ...... (1)

acceleration (a) $= \frac{dv}{dt}$   ......(2)

substituting (2) into (1)

Hence, F $= \frac{mdv}{dt}$

$\frac{dv}{dt} = \frac{F}{m}$

$dv = \frac{F}{m} dt$

$dv = \frac{1}{m}Fdt$

Integrating both sides

$\int\limits {} \, dv = \frac{1}{m} \int\limits {F(t)} \, dt$

The force is acting on the particle between t=0.00 seconds and t=5.00 seconds;

$v = \frac{1}{m} \int\limits^5_0 {F(t)} \, dt$     ......(3)

Substituting the mass (m) =5.00 kg of the particle, equation of the varying force f(t)=6.00t2−4.00t+3.00 and calculating speed at t = 5.00seconds into (3):

$v = \frac{1}{5} \int\limits^5_0 {(6t^{2} - 4t + 3)} \, dt$

$v = \frac{1}{5} |\frac{6t^{3} }{3} - \frac{4t^{2} }{2} + 3t |^{5}_{0}$

$v = \frac{1}{5} |(\frac{6(5)^{3} }{3} - \frac{4(5)^{2} }{2} + 3(5)) - 0|$

$v = \frac{1}{5} |\frac{6(125)}{3} - \frac{4(25)}{2} + 15 |$

$v = \frac{1}{5} |\frac{750}{3} - \frac{100}{2} + 15 |$

$v = \frac{1}{5} | 250 - 50 + 15 |$

$v = \frac{215}{5}$

v = 43 meters per second

The speed v of the particle at t=5.00 seconds = 43 m/s