A force acts on a body of mass 10 kg,resulti | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given,mass $m = 10 \ kg$ and displacement $x = \frac{t^3}{25} \ m$.Velocity $v = \frac{dx}{dt} = \frac{3t^2}{25} \ m/s$.Acceleration $a = \frac{dv

Vedclass · Accepted Answer

$1.152$

Vedclass · Answer

(D) Given,mass $m = 10 \ kg$ and displacement $x = \frac{t^3}{25} \ m$.Velocity $v = \frac{dx}{dt} = \frac{3t^2}{25} \ m/s$.Acceleration $a = \frac{dv}{dt} = \frac{6t}{25} \ m/s^2$.Force $F = m \cdot a = 10 \cdot \left(\frac{6t}{25}\right) = \frac{12t}{5} \ N$.Work done $dW = F \cdot dx = F \cdot \left(\frac{dx}{dt}\right) dt = \left(\frac{12t}{5}\right) \cdot \left(\frac{3t^2}{25}\right) dt = \frac{36t^3}{125} dt$.Total work done in the first $2 \ s$ is $W = \int_{0}^{2} \frac{36t^3}{125} dt = \frac{36}{125} \left[\frac{t^4}{4}\right]_{0}^{2} = \frac{36}{125} \cdot \frac{16}{4} = \frac{36 \cdot 4}{125} = \frac{144}{125} = 1.152 \ J$.

$A$ force acts on a body of mass $10 \ kg$,resulting in its displacement given as $x = \left(\frac{t^3}{25}\right) \ m$,where $t$ is the time in seconds. The work done by the force in the first $2 \ s$ is (in $J$)

Similar Questions

The force on a particle as a function of displacement $x$ (in $x$-direction) is given by $F = 10 + 0.5x$. The work done corresponding to the displacement of the particle from $x = 0$ to $x = 2$ units is:

When a rubber band is stretched by a distance $x$,it exerts a restoring force $F = ax + bx^2$,where $a$ and $b$ are constants. The work done in stretching the rubber band by a distance $L$ from its unstretched position is:

$A$ force is applied to a particle of mass $0.1 \ kg$ as a function of distance as shown in the figure. If it starts from rest at $x = 0$,its velocity at $x = 12 \ m$ is ....... $m/s$.

$A$ block of mass $10\,kg$ is moving along the $x$-axis under the action of a force $F = 5x\,N$. The work done by the force in moving the block from $x = 2\,m$ to $4\,m$ will be ............$J$.

$A$ force $F$ acting on an object varies with distance $x$ as shown in the graph. The force is in $N$ and $x$ is in $m$. The work done by the force in moving the object from $x = 0$ to $x = 6$ is ............. $J$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module