A spherical body of mass $2\, kg$ starting from rest acquires a kinetic energy of $10000\, J$ at the end of $5^{\text {th }}$ second The force acted on the body is ___$N$

Question

A curve is given between the potential energy $U$ of a particle and its position on the $x$-axis as shown. Given: \[ \tan\theta_1 = 1,\qquad \tan\theta_2 = 3,\qquad \tan\theta_3 = -\frac{1}{2} \] If $F_{AB}$ is the force acting on the particle during motion from $A$ to $B$, similarly $F_{BC}$, $F_{CD}$ and $F_{DE}$ are the forces during $B$ to $C$, $C$ to $D$ and $D$ to $E$ respectively, arrange the magnitudes of these forces in decreasing order.

Answer 1

To find the force acting on the spherical body, we need to determine its acceleration. We use the equation for kinetic energy: $ KE = \frac{1}{2}mv^2 $. Here, $ KE = 10000 \, J $ and $ m = 2 \, kg $. Solving for velocity $ v $, we have:

$ 10000 = \frac{1}{2} \times 2 \times v^2 $
$ 10000 = v^2 $
$ v = \sqrt{10000} = 100 \, m/s $

The velocity at the end of the 5th second is $ 100 \, m/s $. Since the body starts from rest, its initial velocity $ u = 0 $. The acceleration $ a $ can be calculated using the equation:

$ v = u + at $
$ 100 = 0 + a \times 5 $
$ a = \frac{100}{5} = 20 \, m/s^2 $

Next, we use Newton's second law to find the force $ F $:

$ F = ma $
$ F = 2 \times 20 = 40 \, N $

The computed force $ F = 40 \, N $ fits within the given range of 40,40, confirming the solution is correct. Thus, the force acting on the body is 40 N.

A spherical body of mass $2\, kg$ starting from rest acquires a kinetic energy of $10000\, J$ at the end of $5^{\text {th }}$ second The force acted on the body is ___$N$

Correct Answer: 40

Solution and Explanation

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