Question

A 15kg block slides on a horizontal surface with an initial speed of 18m/s.The speed of the block is 7m/s after travelling a distance of 250m.How much work done by friction on the block?Calculate the average frictional force acting on the block.What is the coefficient of kinetic friction between the block and the surface?

Answer 1

Answer:

Given,

Initial velocity, u=10ms −1

Displacement, s=50m

Final velocity, v=0

Acceleration of friction, a=μg=10μms ^−2

Apply second kinematic equation of motion,

v ^2 −u ^2 =2as

0−10 ^2 =2(−10μ)×50

μ=0.1

Hence, coefficient of friction is 0.1

Answer 2

Given :

mass of the block (m) = 15 Kg

initial velocity (v$_{1}$) = 7 m/sec

velocity after travelling a distance of 250 m (v$_{2}$)= 7 m/sec

To Find :

(i) How much work done by friction on the block?

(ii) Calculate the average frictional force acting on the block.

(iii) What is the coefficient of kinetic friction between the block and the surface.

Solution :

(i) We know,

            Work done (W) = Change in Kinetic Energy

or,         Work done (W) = $\frac{1}{2} m(v_{1} ^{2} -v_{2} ^{2} )$

or,         Work done (W) = $\frac{1}{2} 15(18 ^{2} -7^{2} )$

or,         Work done (W) = $\frac{1}{2}$ × 15 × (324 - 49)

or,         Work done (W) = $\frac{15}{2}$ × 275

∴            Work done (W) = 2062.5 J

∴ Work done by friction on the block is 2062.5 J

(ii) We know,

  Work done by the friction (W) = Frictional Force(F) × Displacement(ds)

or,                                             W  = F × ds

or,                                    2062.5  = F × 250

or,                                              F  = $\frac{2062.5}{250}$

∴                                                F  = 8.25 N

∴ The frictional force acting on the block is 8.25 N

(iii) We know,

        Frictional Force (F) = Co-efficient of friction (μ) × Normal reaction (N)

or,                                  F  = μ × mg

or,                       8.25       = μ × 15 × 10

or,                           μ        = $\frac{8.25}{150}$

∴                             μ        = 0.055

∴ The coefficient of kinetic friction between the block and the surface (μ) is 0.055 N.