Question

a) A force F=(x÷a-1) is acting on a particle along the x-axis. Determine the work done by the force in moving the particle from x=0 to x=2a.


b) A block of mass 6.0 kg slides from rest at a height of 2.0 m down to a horizontal surface where it passes over a 1.5 m rough patch. After crossing this patch it climbs up another incline which is at an angle of 30° to the ground. The rough patch has a coefficient of kinetic friction . µk=0.25 What height does the block reach on the incline before it comes to rest?

Answer 1

Answer on Question #40130, Physics, Mechanics | Kinematics | Dynamics

a) A force $F = A(x \div a - 1)$ is acting on a particle along the $x$-axis. Determine the work done by the force in moving the particle from $x = 0$ to $x = 2a$.

b) A block of mass $6.0\,\mathrm{kg}$ slides from rest at a height of $2.0\,\mathrm{m}$ down to a horizontal surface where it passes over a $1.5\,\mathrm{m}$ rough patch. After crossing this patch it climbs up another incline which is at an angle of $30{}^{\circ}$ to the ground. The rough patch has a coefficient of kinetic friction. $\mu\mathrm{k} = 0.25$. What height does the block reach on the incline before it comes to rest?

Solution

a) A force acting on a particle along the $x$-axis is

$$F(x) = A\left(\frac{x}{a} - 1\right).$$

The work done by the force $F$ in moving the particle from $x = 0$ to $x = 2a$:

$$W = \int_{0}^{2a} A\left(\frac{x}{a} - 1\right) dx = A\left[\frac{x^2}{2a} - x\right] \left| \begin{array}{c} 2a \\ 0 \end{array} \right. = A\left(\frac{(2a)^2}{2a} - 2a - 0\right) = A \cdot 0 = 0.$$

Answer: 0.

b) $h_1 = 2.0\,\mathrm{m}$ – initial height of block, $h_2$ – final height of block, $m = 6.0\,\mathrm{kg}$ – mass of block, $\mu_{\mathrm{k}} = 0.25$ – coefficient of kinetic friction, $d = 1.5\,\mathrm{m}$ – distance on horizontal rough surface.

According to the law of conservation of energy initial potential energy of block is equal to sum $f$ final potential energy of block and work done against friction force:

$$mgh_1 = mgh_2 + \mu_k mgd.$$

The final height of block:

$$h_2 = h_1 - \mu_k d = 2.0 - 0.25 \cdot 1.5 = 1.6\,\mathrm{m}.$$

Answer: 1.6 m.