Student Exercises 1. A 50-kg box is being pushed along a horizontal surface. The
ID: 1883457 • Letter: S
Question
Student Exercises 1. A 50-kg box is being pushed along a horizontal surface. The coefficient of static friction between the box and the ground is 0.65, and the coefficient of kinetic friction is 0.45. What horizontal force must be exerted on the box for it to accelerate at 1.4 m/s2? 2. The coefficient of static and kinetic friction between a 3.0-kg box and a horizontal desktop are 0.40 and 0.30, respectively. What is the force of friction on the box when a 15-N horizontal push is applied to the box? (Hint: is the box at rest or moving?) 3. An object slides on a level floor. It slows and comes to a stop with a constant acceleration of magnitude 2.4 m/s. What is the coefficient of kinetic friction between the object and the floor? 4. A 55-kg box rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.30, and the coefficient of kinetic friction is 0.20. What horizontal force must be applied to the box to cause it to start sliding along the surface?Explanation / Answer
a. Since the box is accelerating, the net force on the box is the horizontal force minus the kinetic friction force.
f = 0.45
m = 50kg
g = 9.8m/sec2
Kinetic fiction force = 0.45 * 50 * 9.8 = 220.5 N
net force = m.a = 50*1.4 = 70N
Net force = F – 220.5 = 70
F = 220.5 + 70= 290.5 N
hence horizental force is 290.5 N.
2. kinetic friction force = Fk = coefficinet of kinetic friction*m*g = 0.3*3*9.8 =8.82N
Static friction force Fs = 0.4*3*9.8 = 11.76N
hence it is not enough to stop the motion of box as it is less than the force exerted which is 15N
hence net force = 15-8.82 = 6.18N
3.
Force = mass*acceleartion = coefficient of friction * mass* g
hence coeeficient of friction = a/g = 0.24
4. static frictional force = 0.3*55*9.8 = 161.7 N
the horizental force should be 161.7 N so that it starts sliding.