Question

A 4.9 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of an x axis is applied to the block. The force is given by F with arrow(x) = (2.6 − x^2) N, , where x is in meters and the initial position of the block is x=0(a) What is the kinetic energy of the block as it passes through x = 2.1 m?

(b) What is the maximum kinetic energy of the block between x = 0 and x = 2.1 m?

Answer 1

Answer with Explanation:

We are given that

Mass of block=m=4.9 kg

Initial velocity, u=0

$F=(2.6-x^2) i N$

Initial position, x=0

a.We have to find the kinetic energy of the blocks as it passes through x=2.1 m

Work done=Kinetic energy=$\int_{0}^{2.1}(2.6-x^2) dx$

Kinetic energy of the block=$[2.6 x-\frac{x^3}{3}]^{2.1}_{0}$

Kinetic energy of the block=$2.6\times 2.1-\frac{(2.1)^3}{3}-0=2.373 J$

Kinetic energy of the block=2.373 J

b.Initial kinetic energy of block=$K_i=\frac{1}{2}(4.9)(0)=0$

According to work energy theorem

$W=K_f-K_i$

$2.373 =k_f-0$

$k_f=2.373 J$

Hence, the maximum kinetic energy of the block =2.373 J