Question

a) A force F=A[(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 #39588, Physics, Other

a) A force $F = A[\{x / a\} - 1]$ 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 $6.0 \, \mathrm{kg}$ 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)

The basic work relationship $W = Fx$ is a special case which applies only to constant force along a straight line. In the more general case of a force which changes with distance the work may be calculated by performing the integral

$$W = \int_ {x _ {1}} ^ {x _ {2}} F (x) d x = \int_ {0} ^ {2 a} F (x) d x = \int_ {0} ^ {2 a} A [ (x / a) - 1 ] d x$$ $$W = A \int_ {0} ^ {2 a} \left(\frac {x}{a} - 1\right) d x = A \left[ \frac {x ^ {2}}{2 a} - x \right] \bigg | _ {0} ^ {2 a} = A \left[ \frac {(2 a) ^ {2}}{2 a} - 2 a - 0 \right] = A [ 2 a - 2 a ] = 0$$

b)

Given:

$$\mathrm {h} _ {1} = 2. 0 \mathrm {m}$$ $$m = 6. 0 k g$$ $$d = 1. 5 m$$ $$\mu_ {k} = 0. 2 5$$ $$\mathrm {h} _ {1} -?$$

Since the problem involves a change of height and speed, we make use of the Generalized Work-Energy Theorem. Since the block's initial and final speeds are zero, we have

$$W _ {N C} = \Delta E = U _ {f} - U _ {i} = m g h _ {2} - m g h _ {1}$$

The nonconservative force in this problem is friction. To find the work done by friction, we need to know the friction. To find friction, a force, we draw a Free-Body Diagram at the rough surface and use Newton's Second Law.

Along x-axis: $\Sigma F_{x} = m a_{x}, -f_{k} = -m a$

Along y-axis: $\Sigma F_{y} = m a_{y}, N - m g = 0$

The second equation gives $N = m g$ and we know $f_{k} = \mu_{k} N$, so $f_{k} = \mu_{k} m g$. Therefore, the work done by friction is $W_{\text{friction}} = -f_{k} d = -\mu_{k} m g d$. Putting this into equation (1) yields

$$- \mu_{k} m g d = m g h_{2} - m g h_{1}.$$

Solving for $h_{2}$, we find

$$h_{2} = h_{1} - \mu_{k} d = 2.0 - 0.25 \cdot 1.5 = 1.625 \, m.$$

Answer. a) work = 0,

b) $h = 1.625 \, m$.