1
Question
A time dependent force F=6t acts on a particle of mass 1kg. If the particle starts from rest, the work done by the force during the first 1 sec. will be :
A time dependent force F=6t acts on a particle of mass 1kg. If the particle starts from rest, the work done by the force during the first 1 sec. will be :
Open in App
Solution
The correct option is B 4.5J
Work done by time dependent force -
W= \int \vec{F}\cdot \vec{v}\: dt
- wherein
Where \vec{F} and \vec{v} are force and velocity vector at any instant
Force is given F = 6t
F = ma = m\frac{dv}{dt}= 6t
\frac{dv}{dt}=6t(Since m = 1 kg)
dv = 6t dt
On integrating \int_{0}^{v}dV =6\int_{0}^{1}t dt = 3
Therefore v = 3 m/s
Therefore change in kinetic energy in one second = \frac{1}{2}\times m \times 3^{2}-0=4.5 J
Hence,
option (D) is correct answer.
Work done by time dependent force -
W= \int \vec{F}\cdot \vec{v}\: dt
- wherein
Where \vec{F} and \vec{v} are force and velocity vector at any instant
Force is given F = 6t
F = ma = m\frac{dv}{dt}= 6t
\frac{dv}{dt}=6t(Since m = 1 kg)
dv = 6t dt
On integrating \int_{0}^{v}dV =6\int_{0}^{1}t dt = 3
Therefore v = 3 m/s
Therefore change in kinetic energy in one second = \frac{1}{2}\times m \times 3^{2}-0=4.5 J
Hence,
option (D) is correct answer.
Suggest Corrections
0
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
