Question

A block of mass 6.00kg initially at rest is pulled to the right by a constant
horizontal force with magnitude 12.0N. The coefficient of kinetic friction
between the block and the surface is 0.150.
(a) Find the speed of the block after it has moved 3.0m
(b) Suppose the force F is applied at an angle φ. At what angle should the force
applied to achieve the largest possible speed after the block has moved 3.0m
to the right?

Answer 1

Answer on Question#54967 – Physics – Mechanics | Kinematics | Dynamics

$(a)\sqrt{3}\approx 1.732\left(\frac{m}{s}\right)$

$(b)$ atan(0.15) $\approx$ 8.531°

Question

A block of mass $6.00 \, kg$ initially at rest is pulled to the right by a constant horizontal force with magnitude $12.0 \, N$ . The coefficient of kinetic friction between the block and the surface is 0.150.

(a) Find the speed of the block after it has moved $3.0 \, m$

(b) Suppose the force $F$ is applied at an angle $\varphi$ . At what angle should the force applied to achieve the largest possible speed after the block has moved $3.0 \, m$ to the right?

Solution

Denote variables: mass $(m)$ , pulling force $(F_{t})$ , friction force $(F_{f})$ , gravitation force $(F)$ , gravitation acceleration $(g)$ , acceleration of the block $(a)$ , coefficient of kinetic friction $(\mu)$ , velocity $(V)$ , distance $(S)$ , angle regarding surface $(\varphi)$ , total force acting on the block along horizontal direction $(F_{total})$ .

(a) Write down relevant formulae:

$$F _ {f} = \mu F$$ $$m a = F _ {t o t a l}$$ $$F = m g$$ $$F _ {t o t a l} = F _ {t} - F _ {f}$$

Rearrange expressions:

$$m a = F _ {t} - F _ {f}$$ $$m a = F _ {t} - \mu m g$$ $$a = \frac {F _ {t}}{m} - \mu g$$

Integrate both sides with respect to time:

$$V = \left(\frac {F _ {t}}{m} - \mu g\right) t - C _ {1}$$

At initial moment of time block is not moving:

$$C _ {1} = 0; \quad V = \left(\frac {F _ {t}}{m} - \mu g\right) t$$

Integrate with respect to time one more time:

$$S = \left(\frac {F _ {t}}{m} - \mu g\right) \frac {t ^ {2}}{2} + C _ {2}$$

Assume that at initial moment of time block located at the origin:

$$C _ {2} = 0; \quad S = \left(\frac {F _ {t}}{m} - \mu g\right) \frac {t ^ {2}}{2}$$

Factor out $t$ from expression of $S$:

$$S ^ {\frac {1}{2}} = \frac {t \sqrt {2}}{2} \left(\frac {F _ {t}}{m} - \mu g\right) ^ {\frac {1}{2}}; \quad t = \sqrt {2 S} \left(\frac {F _ {t}}{m} - \mu g\right) ^ {- \frac {1}{2}}$$

Plug it in expression of $V$:

$$V = \left(\frac {F _ {t}}{m} - \mu g\right) \sqrt {2 S} \left(\frac {F _ {t}}{m} - \mu g\right) ^ {- \frac {1}{2}}$$ $$V = \sqrt {2 S} \left(\frac {F _ {t}}{m} - \mu g\right) ^ {\frac {1}{2}}$$

Substitute in numbers:

$$V = \sqrt {2 * 3} \left(\frac {1 2}{6} - 0. 1 5 * 1 0\right) ^ {\frac {1}{2}} = \sqrt {6} (2 - 1. 5) ^ {\frac {1}{2}} = \frac {\sqrt {6}}{\sqrt {2}} = \sqrt {3} \left(\frac {m}{s}\right)$$

(b) Write down relevant formulae:

$$F _ {\perp} = F - F _ {t} \sin (\varphi)$$ $$F _ {\parallel} = F _ {t} \cos (\varphi)$$ $$F _ {f} = \mu F _ {\perp}$$ $$F = m g$$ $$m a = F _ {t o t a l}$$ $$F _ {t o t a l} = F _ {\parallel} - F _ {f}$$

Rearrange expressions:

$$m a = F _ {t} \cos (\varphi) - \mu (F - F _ {t} \sin (\varphi))$$ $$m a = F _ {t} \cos (\varphi) - \mu m g + \mu F _ {t} \sin (\varphi)$$ $$a = \frac {F _ {t} \cos (\varphi)}{m} - \mu g + \frac {\mu F _ {t} \sin (\varphi)}{m}$$

Integrate both sides with respect to time:

$$V = \left(\frac {F _ {t} \cos (\varphi)}{m} + \frac {\mu F _ {t} \sin (\varphi)}{m} - \mu g\right) t + C _ {1}$$

At initial moment of time block is not moving:

$$C _ {1} = 0$$

Introduce new variable $\beta (\varphi)$ :

$$\beta = \frac {F _ {t} \cos (\varphi)}{m} + \frac {\mu F _ {t} \sin (\varphi)}{m} - \mu g$$ $$V = \beta t$$

Integrate with respect to time one more time:

$$S = \frac {\beta t ^ {2}}{2} + C _ {2}$$

Assume that at initial moment of time block located at the origin:

$$C _ {2} = 0; \quad S = \frac {\beta t ^ {2}}{2}$$

Factor out $t$ from expression of $S$ :

$$t = \sqrt {\frac {2 S}{\beta}}$$

Plug it in expression of $V$:

$$V = \beta \sqrt{\frac{2S}{\beta}} = \sqrt{2S\beta}$$

Find a maximum of $V$ with respect to $\varphi$ (actually, an extremum, we know that it maximum from physical reasons):

$$\begin{array}{l} \frac{dV}{d\varphi} = \frac{1}{2} \sqrt{\frac{2S}{\beta}} \frac{d\beta}{d\varphi} = \frac{1}{2} \sqrt{\frac{2S}{\beta}} \frac{d\left(\frac{F_t \cos(\varphi)}{m} + \frac{\mu F_t \sin(\varphi)}{m} - \mu g\right)}{d\varphi} = \\ \frac{\sqrt{2}}{2} \sqrt{\frac{S}{\beta}} \frac{F_t}{m} (-\sin(\varphi) + \mu \cos(\varphi)) = 0 \\ \mu \cos(\varphi) = \sin(\varphi) \\ \tan(\varphi) = \mu \\ \varphi = \atan(\mu) \end{array}$$

Plug in numbers:

$$\varphi = \atan(0.15)$$

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