Question:

A $4 \,kg$ object has a velocity, $3.0\, \hat{ i }\, m / s$ at some instant. Eights seconds later, its velocity is $(8.0 \,\hat{ i }+10.0\, \hat{ j }) \,m / s$. Assuming that the object is subjected to a constant net force, the magnitude of the force is

Updated On: Aug 21, 2024
  • $\frac{5 \sqrt{5}}{2} N$
  • $\frac{5 \sqrt{3}}{8} N$
  • $\frac{8 \sqrt{5}}{3} N$
  • $\frac{10 \sqrt{3}}{7} N$
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The Correct Option is A

Solution and Explanation

Given, mass of an object, $m=4 \,kg , u =3 i \,m / s$
$v =(8 \hat{ i }+10 \hat{ j }) m / s$ and time $t=8 \,s$
As we know, equation of the motion
$v = u + a t$
$\Rightarrow \, 8 \hat{ i }+10 \hat{ j }=3 \hat{ i }+ a \times 8$
$\Rightarrow \, a =\frac{1}{8}(5 \hat{ i }+10 \hat{ j }) m / s ^{2}$
So, force $F =m a$
$ F =\frac{4}{8}(\hat{ i }+10 \hat{ j })=\frac{1}{2}(\hat{ i }+10 \hat{ j }) N $
$\Rightarrow\, F=\frac{1}{2} \sqrt{5^{2}+10^{2}}$
$=\frac{5 \sqrt{5}}{2} \,N$
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Concepts Used:

Work-Energy Theorem

The work and kinetic energy principle (also known as the work-energy theorem) asserts that the work done by all forces acting on a particle equals the change in the particle's kinetic energy. By defining the work of the torque and rotational kinetic energy, this definition can be extended to rigid bodies.

The change in kinetic energy KE is equal to the work W done by the net force on a particle is given by,

W = ΔKE = ½ mv2f − ½ mv2i

Where, 

vi → Speeds of the particle before the application of force

vf → Speeds of the particle after the application of force

m → Particle’s mass

Note: Energy and Momentum are related by, E = p2 / 2m.