Question

A 10-kilogram block rests initially on a table as shown in cases I and II. The coefficient of sliding friction between the block and the table is 0.2. The block is connected to a cord of negligible mass, which hangs over a mass less friction less pulley. In case I a force of 50 newtons is applied to the cord. In case II an object of mass 5 kilograms is hung on the bottom of the cord. Use g = 10 meters per second squared.

a. Calculate the acceleration of the 10-kilogram block in case I. 

b. On the diagrams below, draw and label all the forces acting on each block in case II

c. Calculate the acceleration of the 10-kilogram block in case II.

Answer 1

a. ΣF = ma; 50 N – f = ma where f = µN and N = mg gives 50 N – µmg = ma; a = 3 m/s²

b.

c. ΣF = ma for each block gives W₅ – T = m₅a and T – f = m₁₀a. Adding the two equations gives W₅ – f = (m₅ + m₁₀)a, or a = 2m/s²

Answer 2

(a) 5 points

From direct application of Newton's second law, letting f denote the frictional force on the block

50 - f - ma

N - mg - 0

From the definition of the frictional force

f = μN

Eliminating f and N, one obtains

50 - μmg = ma or

a = 3m/s²

(b) 

(c) 4 points

Again, from Newton's second law one has

m₅a = W₅ - T

m₁₀a = T - f

Eliminating the tension T

(m₅ + m₁₀)a = W₅ - f or

a = 2 m/s²