Question

A block of mass 2.20 kg is accelerated across a rough surface by a rope passing over a pulley, as shown below. The tension in the rope is 10.0 N, and the pulley is 10.0 cm above the top of the block. The coefficient of kinetic friction is 0.400. a. Determine the acceleration of the block when x = 0.400 m. b. Find the value of x at which the acceleration becomes zero.

Answer 1

(a)
tan(θ) = 0.4/0.1
θ = 76

Friction Force = u* N
Where N = m*g - T*sin(90-76)

T*cos(90-76) - Friction Force = Mass * Acceleration
T * cos(14) - u* (m*g - T*sin(14)) = m*a
10.0 * cos(14) - 0.4* (2.20*9.8 - 10.0*sin(14)) = 2.20*a
Solving for a
a = 0.93 m/s^2

Acceleration, a = 0.93 m/s^2

(b)
For Acceleartion to be zero -
T*cos(90 - θ) - Friction Force = 0
T* cos(90 -θ) = u* (m*g - T*sin(90 -θ))
T * sin(θ) = u* (m*g - T*cos(θ))
10.0 *sin(θ) = 0.4 * (2.2 * 9.8 - 10.0 *cos(θ) )
10.0 *sin(θ) + 4 * cos(θ) = 8.624

sin(θ) = x/sqrt(0.1^2 + x^2)
cos(θ) = 0.1/sqrt(0.1^2 + x^2)

10.0 *(x/sqrt(0.1^2 + x^2)) + 4*(0.1/sqrt(0.1^2 + x^2)) = 8.624

Solving for x
x = 0.061 m