A 43.3 kg wagon is towed up a hill inclined at 16.8◦ with respect to the horizontal.?
Update:
The tow rope is parallel to the incline and has a tension of 143 N in it. Assume that the wagon
starts from rest at the bottom of the hill, and neglect friction.
The acceleration of gravity is 9.8 m/s^2.
How fast is the wagon going after moving 66.4 m up the hill?
Answer in units of m/s.
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Answers & Comments
Verified answer
sin 16.8 = 0.2890
Two different approaches :
a)
opponent force Fo = m*g*sin 16.8 = 43.3*9.806*0.2890 = 122.71 N
acceleration a = (F-Fo)/m = (143-122.71)/43.3 = 0.4686 m/sec^2
66.4*2 = a*t^2
t = √132.8/0.4686 = 16.835 sec
Vfin = a*t = 16.835*0.4686 = 7.889 m/sec
b)
m*g*ℓ*sin 16.8 + m/2*Vfin^2 = F*ℓ
Vfin = √(66.4*(143-43.3*9.806*0.2890)*2/43.3 = 7.889 m/sec
F(net) = F(applied) - F(weight component downwards along incline)
ma = 143 - mg sin16.8
a = (143 - (43.3*9.8*sin16.8))/(43.3)
we will avoid calculating to reduce rounding errors.
v² = u² + 2as
since wagon starts form rest u = 0
v = √2as
v = √{(2*66.4[(143 - (43.3*9.8*sin16.8))/(43.3)]}
v = 7.9 m/s
Kinematic solution
a = F/m
a = (143 - 43.3(9.8)sin16.8) / 43.3
a = 0.470 m/s²
v² = u² + 2as
v² = 0² + 2(0.470)(66.4)
v² = 62.4198...
v = 7.90 m/s
Energy solution
The work of the pull will convert to increased gravity potential and kinetic energies.
PE + KE = U
mgh + ½mv² = Fd
v² = (2/m)(Fd - mgh)
v² = (2/m)(Fd - mgdsinθ)
v² = (2d)(F/m - gsinθ)
v² = (2(66.4)(143 / 43.3 - 9.8sin16.8)
v² = 62.4198...
v = 7.90 m/s
I hope this helps.
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