Question #219021

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?



1
Expert's answer
2021-07-20T09:46:21-0400

1.1.

F=maF = ma

F1=430N;Ffr=110NF_1 = 430N;F_{fr}=110N

F=F1cos35°Ffr=242.23NF = F_1\cos35\degree-F_{fr}=242.23N

m=m1+m2=11+30=41kgm = m_1+m_2= 11+30=41kg

a=Fm=242.2341=5.9ms2a = \frac{F}{m}=\frac{242.23}{41}=5.9\frac{m}{s^2}

Answer: a=5.9ms2\text{Answer: }a = 5.9\frac{m}{s^2}

2.2.

g=Gmer2g= G\frac{m_e}{r^2}

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

G=6.673841011G= 6.67384*10^{-11}

me=5.97221024kgm_e= 5.9722*10^{24}kg

r=6.371106mr = 6.371*10^6m

h=60km=60000mh =60km = 60000m

g=6.6738410115.97221024(6.371106+60000)=9.64ms2g'= 6.67384*10^{-11}*\frac{ 5.9722*10^{24}}{(6.371*10^6+60000)}=9.64\frac{m}{s^2}

Answer: g=9.64ms2\text{Answer: } g'=9.64\frac{m}{s^2}

3.3.

T=PT = |P|

P=m(g+a)P= m(g+a)

a1=2.2ms2a_1 = -2.2\frac{m}{s}^2

T1=5.5(9.82.2)=41.8NT_1= 5.5*(9.8-2.2)= 41.8N

a2=1.7ms2a_2= 1.7\frac{m}{s}^2

T2=5.5(9.8+1.7)=63.25NT_2 = 5.5(9.8+1.7)=63.25N

Answer: T1=41.8N;T2=63.25N\text{Answer: }T_1= 41.8N;T_2 =63.25N

4.4.

F=ma=272.82=76.14F = ma= 27*2.82=76.14

F=F1FfrF =F_1-F_{fr}

Ffr=F1F=11076.14=33.86NF_{fr} = F_1-F = 110-76.14= 33.86N

μ=FfrN=Ffrmg=33.86279.8=0.13\mu=\frac{F_{fr}}{N}=\frac{F_{fr}}{mg}=\frac{33.86}{27*9.8}=0.13

Answer: μ=0.13\text{Answer: }\mu=0.13

5.5.

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

a=4.73ms2a = 4.73\frac{m}{s^2}

Answer: a=4.73ms2\text{Answer: }a = 4.73\frac{m}{s^2}


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