Please help ~ Two blocks are arranged at the ends of a massless string as 3.21 kg (box hanging from table)
4.96 kg (box on table). The system starts from rest. When the 3.21 kg mass has fallen through 0.4 m, its downward speed is 1.27 m/s. The acceleration of gravity is 9.8 m/s2 .
What is the frictional force between the 4.96 kg mass and the table?
Answer in units of N.
CR, check your 2-20-11,11:27pm post.
To find the frictional force between the 4.96 kg mass and the table, we need to first calculate the tension in the string.
Starting from rest, the 3.21 kg mass falls through a distance of 0.4 m with a downward speed of 1.27 m/s. Using this information, we can find the final velocity of the 3.21 kg mass just before it hits the ground.
We can use the kinematic equation:
v² = u² + 2as
Where:
v = final velocity (1.27 m/s)
u = initial velocity (0 m/s, since the block starts from rest)
a = acceleration (acceleration due to gravity = 9.8 m/s²)
s = distance fallen (0.4 m)
Rearranging the equation, we have:
1.27² = 0 + 2 * 9.8 * 0.4
1.6129 = 7.84 * 0.8
1.6129 = 6.272
Now, we have the final velocity of the 3.21 kg mass just before it hits the ground, which is 1.6129 m/s. This will also be the final velocity of the 4.96 kg mass because they are connected by the string.
Since the 4.96 kg mass is on a horizontal table, there is no vertical motion. Therefore, the net force acting on it will be zero in the vertical direction. Since the only vertical forces acting on the mass are its weight and the tension in the string, the tension in the string will be equal to the weight of the mass.
Using the equation:
Weight = mass * acceleration due to gravity
Tension in the string = 4.96 kg * 9.8 m/s²
Tension in the string = 48.608 N
Now, we know the tension in the string is 48.608 N, and this is also the force acting on the 4.96 kg mass due to the friction with the table. Therefore, the frictional force between the 4.96 kg mass and the table is 48.608 N.