Work Done By Friction In Stopping A Moving Object
Strings (SiPjAjk) = S7P3A31 Base Sequence = 12735 String Sequence = 12735 - 3 - 31
A luggage weighing 490 lbs is being moved on a horizontal conveyor belt at 6 miles per hour. The conveyor is then brought to an abrupt stop. The luggage then slided for 1.37 secs during which time it covered 6 feet before coming to a complete stop.
(a) Calculate the work done by friction in bringing the luggage to a complete stop. The coefficient of friction between conveyor belt and luggage is 0.2.
(b) Confirm that the distance covered during the slide and the duration of the slide are 6 feet and 1.37 secs respectively.
(a) S7P3A31 (Force - Pull).
Pj Problem of interest is of type force. Work problems are generally of type force because force is the doer of work. In the case of white-collar work, intellectual force actuates materiality.
Consider the above diagram (FW.1):
Let W be the work done by friction to stop the moving object.
When conveyor belt is stopped abruptly, the only force acting to bring the motion of the luggage to a stop is the friction force between the luggage and the belt. This friction force is equal to the weight of the luggage. The work done by friction is equal to the kinetic energy of luggage before conveyor belt was abruptly stopped.
Speed of conveyor belt = 6 miles/hr = 8.8 ft/sec
Mass of luggage = 490/32 = 15.31 lbs
So, Kinetic Energy = (mv2)/2 = (15.31 x 8.8 x 8.8)/2 = 592.8 ft-lbs.
So, W = kinetic energy = 592.8 ft-lbs.
(b) Strings are: S7P4A41 (Linear motion) and S7P5A51 (Physical Change - Duration) respectively.
To calculate distance assuming it was not given, we equate kinetic energy with work done by friction
So, 592.8 = (friction force) x distance
So, 592.8 = μ490 x distance = 0.2 x 490 x distance
So, distanced luggage slided = 592.8/98 = 6 feet.
To calculate duration of slide assuming it was not given, we use the impulse momentum formula:
R x t = m(vf - vo).
R is the resultant force, t is time the resultant force is acting on the luggage, m is mass of the luggage, vf is the luggage's final velocity and vo is the luggage's original velocity.
So, R = friction force = -98 lbs; m = 490/32; vo = 8.8 ft/sec; vf = 0.
So, -98t = (490/32)(0 - 8.8)
So, t = 1.37 secs.
In SI units: mass is in kilograms, distance is in meters, velocity is in meters/sec, acceleration is in meters/sec2, force is in Newton and work is in Joules.
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