initially, a 10kg block is sliding along a level floor at a speed of 5 m/s. the
ID: 1394164 • Letter: I
Question
initially, a 10kg block is sliding along a level floor at a speed of 5 m/s. the coefficient of kinetic friction between the block and floor is 0.5. a. what are the initial kinetic, potential, and total energies of the block? b. what is the force of friction by the table on the block? c. what is the work done by the frictional force as the block slides 2 m across the floor? d. what are the kinetic, potential, and total energies of the block after sliding 2 m? e. what is the blocks speed after sliding 2m?
Explanation / Answer
Here ,
mass , m =10 Kg
speed , u = 5 m/s
uk = 0.5
a) initial kinetic energy = 0.5 * m * u^2
initial kinetic energy = 0.5 * 10 * 5^2
initial kinetic energy = 125 J
the initial kinetic energy is 125 J
potential energy = 0 J
total energy = 125 + 0
total initial energy = 125 J
b)
normal reaction , N = mg
N = 10 * 9.8
friction force , f = 0.5*10*9.8
f = 49 N
frictional force is 49 N
c)
work done by friction = - friction * distance
work done by friction = -49 * 2
work done by friction = -98 J
d)
using work energy theorum ,
kinetic energy = 125 - 98
kinetic energy = 27 J
the kinetic energy of block is 27 J
potential energy = 0 J
total energy = 27 J
e)
kinetic energy = 0.5 mv^2
27 = 0.5 * 10 * v^2
v = 2.32 m/s
the speed of block is 2.32 m/s
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