Answer to Question #219021 in Classical Mechanics for Laura

Question

1. A student is pulling an 11.0 kg sled along flat ground. The student's 30.0 kg sister then sits in the sled. The student then applies a 430 N force at an angle of 35° N of E. If the force of friction is 180 N, what is the acceleration of the sister sitting in the sled.

2. What is the gravitational field strength at a distance of 60.0 km above the surface of the Earth?

3. A 5.5 kg bowling ball hangs by a thin wire in an elevator. Calculate the tension in the wire supporting the bowling ball if the elevator moves: Downward with a constant acceleration of 2.2 m/s2. Upward with a constant acceleration of 1.7 m/s2.

4. A 27.0 kg object is being pulled east across the ground with a horizontal force of 110 N. If the acceleration of the box is 2.82 m/s2, what is the coefficient of kinetic friction?

5. A 6.0 kg mass is placed on a 20° incline which has a coefficient of friction of 0.15. What is the acceleration of the mass down the incline?

Answer 1

$1.$

$F = ma$

$F_1 = 430N;F_{fr}=110N$

$F = F_1\cos35\degree-F_{fr}=242.23N$

$m = m_1+m_2= 11+30=41kg$

$a = \frac{F}{m}=\frac{242.23}{41}=5.9\frac{m}{s^2}$

$\text{Answer: }a = 5.9\frac{m}{s^2}$

$2.$

$g= G\frac{m_e}{r^2}$

$g'= G\frac{m_e}{(r+h)^2}$

$G= 6.67384*10^{-11}$

$m_e= 5.9722*10^{24}kg$

$r = 6.371*10^6m$

$h =60km = 60000m$

$g'= 6.67384*10^{-11}*\frac{ 5.9722*10^{24}}{(6.371*10^6+60000)}=9.64\frac{m}{s^2}$

$\text{Answer: } g'=9.64\frac{m}{s^2}$

$3.$

$T = |P|$

$P= m(g+a)$

$a_1 = -2.2\frac{m}{s}^2$

$T_1= 5.5*(9.8-2.2)= 41.8N$

$a_2= 1.7\frac{m}{s}^2$

$T_2 = 5.5(9.8+1.7)=63.25N$

$\text{Answer: }T_1= 41.8N;T_2 =63.25N$

$4.$

$F = ma= 27*2.82=76.14$

$F =F_1-F_{fr}$

$F_{fr} = F_1-F = 110-76.14= 33.86N$

$\mu=\frac{F_{fr}}{N}=\frac{F_{fr}}{mg}=\frac{33.86}{27*9.8}=0.13$

$\text{Answer: }\mu=0.13$

$5.$

$a = g(\sin\alpha+\mu\cos\alpha)= 9.8(\sin20\degree+0.15*\cos20\degree)$

$a = 4.73\frac{m}{s^2}$

$\text{Answer: }a = 4.73\frac{m}{s^2}$