Question

2. A 5.0 kg box is moving on a horizontal surface at 0.50 m/s with a rope applying a constant tension on the box of 100 N at an angle of 10° above the horizontal (above the direction of motion). The coefficient of kinetic friction between the box and the surface is 0.25. What is the speed of the box after it has moved 25 cm?

Answer 1

here,

the mass of box, m = 5 kg

the initial horizontal velocity , u = 0.5 m/s

the constant tension , T = 100 N

theta = 10 degree

coefficient of kinetic friction , uk = 0.25

the acceleration of box , a = ( net force)/effctive mass

a =( T * coa(theta) - uk * ( m * g - T * sin(theta)) ) /m

a = ( 100 * cos(10) - 0.25 * ( 5 * 9.81 - 100 * sin(10)) ) /5 m/s^2

a = 18.1 m/s^2

s = 25 cm = 0.25 m

using third equation of motion

v^2 - u^2 = 2 * a * s

v^2 - 0.5^2 = 2 * 18.1 * 0.25

solving for v

v = 3.05 m/s

the final velocity of box is 3.05 m/s