# Question #3bb16

Mar 31, 2017

If you mean the last problem, the answer is 3) 3.92 m/s^2

#### Explanation:

Forces: ${m}_{B} g$
acting on two masses without friction
so
${m}_{B} g = \left({m}_{A} + {m}_{B}\right) a$
$a = {m}_{B} / \left({m}_{A} + {m}_{B}\right) g = 3.92$ $\frac{m}{s} ^ 2$

Mar 31, 2017

#### Explanation:

Constant velocity means zero acceleration i.e. sum of forces is $T \cos 30 = \mu \left(m g - T \sin 30\right)$
or $\mu = 5 \frac{\sqrt{3}}{2} / \left[\left(3 \cdot 9.81\right) - 2.5\right] = 0.161$

Apr 1, 2017

Q no 11 $\to$option (2)

#### Explanation:

Initial velocity of the car $u = 0$

Applied force $F = 400 N$

Final velocity of the car $v = 14 k m \text{/"h=(14xx1000)/3600=35/9m"/} s$

Time taken to change in velocity $t = 12 s$

Acceleration produced $a = \frac{v - u}{t} = \frac{35}{12 \times 9} = \frac{35}{108} m \text{/} {s}^{2}$

So by Newton's law the mass of the car

$M = \frac{F}{a} = \frac{400}{\frac{35}{108}} k g = \frac{400 \times 108}{35} k g \approx 1230 k g$(rounding up to 3 significant figure)

Apr 1, 2017

Q No 12$\to$ Option (1) 69.6N

#### Explanation:

Considering the equilibrium of forces in the vertical horizontal direction we can write

${T}_{1} \sin 40 + {T}_{2} \sin 45 = 10 \times 9.81 = 98.1 N \ldots \ldots . \left[1\right]$

${T}_{1} \cos 40 = {T}_{2} \cos 45. \ldots . \left[2\right]$

Adding [1] and [2] we get

${T}_{1} \sin 40 + {T}_{2} \sin 45 + {T}_{1} \cos 40 = 98.1 + {T}_{2} \cos 45$

$\implies {T}_{1} \sin 40 + \cancel{{T}_{2} / \sqrt{2}} + {T}_{1} \cos 40 = 98.1 + \cancel{{T}_{2} / \sqrt{2}}$

$\implies {T}_{1} = \frac{9.81}{s n 40 + \cos 49} \approx 69.6 N$