Energy problem with a block Initially, a 10 kg block is sliding along a level fl
ID: 1344012 • Letter: E
Question
Energy problem with a block
Initially, a 10 kg block is sliding along a level floor at a speed of 5 m/s. The coefficient of kinetic friction between the block and the floor is 0. 5. What are the initial kinetic, potential, and total energies of the block? What is the force of friction by the table on the block? What is the work done by the frictional force as the block slides 2. 0 m across the floor (i. e. . how much energy is removed from the system)? What are the kinetic, potential, and total energies of the block after sliding What is the block's speed after sliding 2. 0 m?Explanation / Answer
here,
mass of the block , m = 10 kg
initial speed , v0 = 5 m/s
uk = 0.5
(a)
initial kinetic energy , KEi = 0.5 * m*v0^2
KEi = 125 J
the initial kinetic energy is 125 J
as it is on floar level
the potential energy is 0
the total energy = kinetic energy
the total energy is 125 J
(b)
the friction force , ff = uk*m*g
ff = 49 N
the friction force acting is 49 N
(c)
the work done by the friction force , Wf = - ff* d* cos(180)
Wf = - 98 J
the work done by the friction force is - 98 J
(d)
final velocity be v
accelration , a = uk*g
a = - 4.9 m/s^2
using third equation of motion
v^2 - v0^2 = 2*a*s
v^2 - 5^2 = - 2*4.9 * 2
v = 2.32 m/s
the final kinetic energy , KEf = 0.5 * m* v^2
KEf = 27 J
as it is on floar level
the potential energy is 0
the total energy = kinetic energy
the total energy is 27 J
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