Question

A block of mass m = 98 kg slides along a horizontal surface. The coefficient of friction between the block and the surface is μ_k = 0.38. The block has an initial speed of v_o = 13 m/s in the positive x-direction as shown.

a) write an expression for x-component of the frictional force the block experiences, F(f), in terms of the given variables and variables available in the palette

b) what is the magnitude of the frictional force in N?

c) How far will the block travel, in meters, before coming to rest?

Answer 1

a)

N = Normal force on the block

m = mass of the block = 98 kg

Weight of the block is given as

W = mg

W = (98) (9.8)

W = 960.4 N

Along the vertical direction, normal force on the block balances the weight of the block, hence

N = W

N = 960.4 N

f_k = x- component of frictional force the block experience

_k = Coefficient of kinetic friction = 0.38

x- component of frictional force the block experience is given as

f_k = - _k N

f_k = - _k W

f_k = - _k mg (the negative sign indicates the opposite direction of the frictional force relative to direction of motion)

b)

magnitude of the frictional force is given as

f_k = _k mg

inserting the values

f_k = (0.38) (98) (9.8)

f_k = 365 N

c)

v_o = initial speed of the block = 13 m/s

v = final speed of the block = 0 m/s

d = distance traveled before stopping

Using work-change in kinetic energy theorem

Work done by friction = Change in kinetic energy

- f_k d = (0.5) m (v_f² - v_o²)

- (365) d = (0.5) (98) (0² - 13²)

d = 22.7 m