Home
Class 12
PHYSICS
Under the action of a force a 2 kg body ...

Under the action of a force a 2 kg body moves such that its position x in meters as a function of time t is given by `x=(t^(4))/(4)+3.` Then work done by the force in first two seconds is

A

6 J

B

10 J

C

7 J

D

64 J

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of calculating the work done by the force on a 2 kg body moving according to the position function \( x(t) = \frac{t^4}{4} + 3 \), we will follow these steps: ### Step 1: Find the velocity as a function of time The velocity \( v(t) \) is the derivative of the position function \( x(t) \) with respect to time \( t \). \[ v(t) = \frac{dx}{dt} = \frac{d}{dt} \left( \frac{t^4}{4} + 3 \right) \] Differentiating, we get: \[ v(t) = \frac{4t^3}{4} + 0 = t^3 \] ### Step 2: Calculate the velocity at \( t = 2 \) seconds Now, we will find the velocity of the body at \( t = 2 \) seconds. \[ v(2) = (2)^3 = 8 \, \text{m/s} \] ### Step 3: Calculate the initial kinetic energy at \( t = 0 \) seconds At \( t = 0 \), the velocity is: \[ v(0) = (0)^3 = 0 \, \text{m/s} \] The initial kinetic energy \( KE_0 \) is given by: \[ KE_0 = \frac{1}{2} mv(0)^2 = \frac{1}{2} \times 2 \, \text{kg} \times (0)^2 = 0 \, \text{J} \] ### Step 4: Calculate the kinetic energy at \( t = 2 \) seconds Now, we will calculate the kinetic energy \( KE \) at \( t = 2 \) seconds. \[ KE = \frac{1}{2} mv(2)^2 = \frac{1}{2} \times 2 \, \text{kg} \times (8 \, \text{m/s})^2 \] Calculating this gives: \[ KE = \frac{1}{2} \times 2 \times 64 = 64 \, \text{J} \] ### Step 5: Calculate the work done by the force The work done \( W \) by the force is equal to the change in kinetic energy: \[ W = KE - KE_0 = 64 \, \text{J} - 0 \, \text{J} = 64 \, \text{J} \] Thus, the work done by the force in the first 2 seconds is: \[ \boxed{64 \, \text{J}} \]
Promotional Banner

Similar Questions

Explore conceptually related problems

Under the action of a force, a 2 kg body moves such that its position x as a function of time is given by x =(t^(3))/(3) where x is in meter and t in second. The work done by the force in the first two seconds is .

Under the action of a force, a 2 kg body moves such that its position x as a function of time is given by x =(t^(3))/(3) where x is in metre and t in second. The work done by the force in the first two seconds is .

Under the action of foece, 1 kg body moves such that its position x as a function of time t is given by x=(t^(3))/(3), x is meter. Calculate the work done (in joules) by the force in first 2 seconds.

Under the action of a force, a 1 kg body moves, such that its position x as function of time t is given by x=(t^(3))/(2). where x is in meter and t is in second. The work done by the force in fiest 3 second is

A force acts on a 2 kg object so that its position is given as a function of time as x=3t^(2)+5. What is the work done by this force in first 5 seconds ?

The displacement x of a body of mass 1 kg on a horizontal smooth surface as a function of time t is given by x = (t^4)/4 . The work done in the first second is

The position of a particle along x-axis at time t is given by x=1 + t-t^2 . The distance travelled by the particle in first 2 seconds is

A body of mass 3 kg is under a force , which causes a displacement in it is given by S = (t^(3))/(3) (in metres). Find the work done by the force in first 2 seconds.

A particle of mass 2 kg starts motion at time t = 0 under the action of variable force F = 4t (where F is in N and t is in s). The work done by this force in first two second is

A point moves such that its displacement as a function of time is given by x^3 = t^3 + 1 . Its acceleration as a function of time t will be