Question:
Suppose a hanging 1.0kg lab mass is attached to a 4.0 kg block on the table,
a) if the coefficient of kinetic friction is .2, what is the acceleration of the block?
b) what would be the coefficient of friction of static friction in order for the block to remain motionless?
a) if the coefficient of kinetic friction is .2, what is the acceleration of the block?
b) what would be the coefficient of friction of static friction in order for the block to remain motionless?
The normal is the applied force on the block to counteract the weight force. Friction is the frictional force on the surface of the table. T is the force that the small block has. If the N and W are balanced the block remains motionless in the vertical direction.
For the object to remain constant, the weight and normal must be equal. The mass of both blocks combined is 5 kg. Gravity acts upon both blocks. The weight force equals mass times the gravitational constant.
W = ma
W = (5kg)(9.8)
W = 39.2 N
The weight force is 39.2 N.
W = ma
W = (5kg)(9.8)
W = 39.2 N
The weight force is 39.2 N.
a) Friction = μN The frictional force was determined by multiplying the coefficient of kinetic friction by the normal
= .2(39.20 force. The coefficient of kinetic friction was stated in the problem to be .2. The normal is equal to
= 7.84 the weight if the block is motionless in the vertical direction and weight was determined earlier to
be 39.2 N.
F=ma Net force = mass x acceleration was used to find acceleration. The two forces in the horizontal
9.8-7.84 = 5a direction are the force the small block has on the big block and the frictional force provided by the
5a = 1.96 surface. The only force on the small block is the weight force which equals mass x acceleration.
a = .392 m/s2 mass = 1kg
a = 9.8 m/s2
1 x 9.8 = 9.8 N
This is the tension force. The other force is the frictional force which we found to be 7.84 N.
The difference of these two values equals mass x acceleration. The mass of both combined is 5.
Solving for a, we find it to be .392 m/s2
b) 9.8 = μN In order for the block to remain motionless, the tension force and frictional force must be equal.
9.8 = μ39.2 We know the tension force is 9.8 and the normal is 39.2, we plug it into Friction = μN and figure out
μ = .25 that the coefficient of friction is .25.
= .2(39.20 force. The coefficient of kinetic friction was stated in the problem to be .2. The normal is equal to
= 7.84 the weight if the block is motionless in the vertical direction and weight was determined earlier to
be 39.2 N.
F=ma Net force = mass x acceleration was used to find acceleration. The two forces in the horizontal
9.8-7.84 = 5a direction are the force the small block has on the big block and the frictional force provided by the
5a = 1.96 surface. The only force on the small block is the weight force which equals mass x acceleration.
a = .392 m/s2 mass = 1kg
a = 9.8 m/s2
1 x 9.8 = 9.8 N
This is the tension force. The other force is the frictional force which we found to be 7.84 N.
The difference of these two values equals mass x acceleration. The mass of both combined is 5.
Solving for a, we find it to be .392 m/s2
b) 9.8 = μN In order for the block to remain motionless, the tension force and frictional force must be equal.
9.8 = μ39.2 We know the tension force is 9.8 and the normal is 39.2, we plug it into Friction = μN and figure out
μ = .25 that the coefficient of friction is .25.