Question

6. A 2.0 kg block slides to the right across a horizontal surface. Let's define a coordinate system where the +x-direction is to the right. At x=0, the block's velocity is +5.0 m / s (to the right). The magnitude of the kinetic friction force between the block and surface is 10N.

a. Draw a force diagram of the block. Which forces expert zero work on the block? Explain why.

c. What is the work that the kinetic friction force exerts on the block as the block slides from x=0 to x=0.9 m ? Does this transfer energy to the block or transfer energy from the block? Explain.

d. What is the kinetic energy of the block at x=0.9 m ? What is the block's speed at the point?

Answer 1

(a)

N and G exert 0 work on the block because they are oriented perpendicular to the trajectory (cos90=0).

(b)

$E_{k}=\frac{mv^{2}}{2}$

E_k=25J

(c)

W=-10*0.9=-9J

(minus is due to the fact that the sense of the friction force is opposite to that of the sense of the movement)

None of them. This energy is 'lost' in interaction forces between the surface and the bloc, usually as heat.

(d)

$\Delta E_{k}=W\Leftrightarrow E_{kf}-E_{ki}=W\Rightarrow E_{kf}=E_{ki}+W\: \left ( 1 \right )$

E_kf=25-9=16 J

$v_{f}=\sqrt{\frac{2E_{kf}}{m}}$

v_f=4m/s

(e)

Eq.(1) gives: $\Delta E_{k}=W_{f}\Leftrightarrow E'_{kf}-E_{ki}=W\Rightarrow E'_{kf}=E_{ki}+W_{_{f}}\: \left ( 1 \right )$

$E'_{kf}$=0 (stop condition) --> W_f=-E_ki

$W_{f}=F_{f}d_{s}\Rightarrow d_{s}=\frac{W_{f}}{F_{f}}$

Thus, $d_{s}=\frac{-E_{ki}}{F_{f}}$

d_s=25/10=2.5m

Wf=2.5*10=25 J