A nasty person puts a man m=60 kg into a cart m=100 kg, and places the cart agai
ID: 1261908 • Letter: A
Question
A nasty person puts a man m=60 kg into a cart m=100 kg, and places the cart against a large spring with k=9000 N/m, compresses the spring by d=1.75 m, then releases the cart. The man and cart shoot down a horizontal track with coefficient of friction 0.10. The track ends at the edge of an extremely deep abyss which is 4.0 m wide. The other side of the abyss is higher than the side of the track by 5.0 m. Luckily for the man, there is a 4.0 m long cable suspended a horizontal distance of 10.0 m from the compressed spring and 2.0 m from the edge of the cliff that the man can grab onto. Assume the freed end of the cable is at ground level, as is the man initially. The man wishes to get to the other side of the abyss, and releases the steel cable when the tension in the cable is 3397 N. Using g=10 m/s^2, coefficient of energy equations, and circular motion; will the man make it to the other side safely?
A nasty person puts a man m=60 kg into a cart m=100 kg, and places the cart against a large spring with k=9000 N/m, compresses the spring by d=1.75 m, then releases the cart. The man and cart shoot down a horizontal track with coefficient of friction 0.10. The track ends at the edge of an extremely deep abyss which is 4.0 m wide. The other side of the abyss is higher than the side of the track by 5.0 m. Luckily for the man, there is a 4.0 m long cable suspended a horizontal distance of 10.0 m from the compressed spring and 2.0 m from the edge of the cliff that the man can grab onto. Assume the freed end of the cable is at ground level, as is the man initially. The man wishes to get to the other side of the abyss, and releases the steel cable when the tension in the cable is 3397 N. Using g=10 m/s^2, coefficient of energy equations, and circular motion; will the man make it to the other side safely?Explanation / Answer
PE of compressed spring = (1/2)*k*x^2 = 0.5*9000* 1.75^2 = 13781 Jules
work done against friction = friction force * distance moved by cart and man
friction force = (mass of car + mass of man )* g * coefficient of friction = (100 + 60) * 9.8 * 0.1 = 156.8 N
work done against friction = friction force * distance moved by cart and man
work done against friction = 156.8 * 10 = 1568 Jules
the energy carried by the man and cart just before he holds the suspended rope = PE of compressed spring - work done against friction
the energy carried by the man and cart just before he holds the suspended rope = 13781 - 1568 = 12213 Jules
the energy carried by the man and cart just before he holds the suspended rope = (1/2)*m(cart*man)*v^2
v = 12.3557 m/s
let the man leaves the rope when the rope is making angle ( theta + 90 )with vertical
therefore,
the total vertical movement done by the man in circular motion is = 4 + 4 sin (theta)
KE lost due to this vertical lift of the man = mgh = 160 * 9.8* 4 *( 1 + sin(theta) )
the energy left at this vertical hight = the energy of the man when he holds the rope - KE lost due to this vertical lift of the man
the energy left at this vertical hight = 4579.9 - 160 * 9.8* 4 *( 1 + sin(theta) )
the tension at this point is given as T = 33397 N
according to force concept
T + mg sin(theta) = m* v(at the time when he leavs the rope) ^2 / r
the velocity at the time when he leavs the rope = v
by energy conservation
(1/2)*m*v^2 = the energy left at this vertical hight = 12213 - 60 * 9.8* 4 *( 1 + sin(theta) )
v^2 = [12213 - 60 * 9.8* 4 *( 1 + sin(theta) ) ] *2 / 160
therefore
puttinv v^2 value in
T + 160g sin(theta) = 160* ([[12213 - 60 * 9.8* 4 *( 1 + sin(theta) ) ] *2 / 160] *2 / 60) / 4
therefore,
theta = -ve value
which is not possibl eas per assumption of our case so... he will never make up to this top of cliff which is 5 meter high
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