Question

an experiment is conducted in which a spring scale is attached to a block of iron with a weight of 82 n at rest on a sheet of aluminum. an applied force of 64 n is required to cause the block to move initially and 40 n is required to pull the block forward at a constant speed once it is in motion. which are the coefficients of static and kinetic friction? (1 point)

1.3 and 2.1

0.49 and 0.78

2.1 and 1.3

0.78 and 0.49

Explanation:

Step1: Recall static - friction formula

The formula for static friction is $F_{s}=\mu_{s}N$, where $F_{s}$ is the force of static friction, $\mu_{s}$ is the coefficient of static friction, and $N$ is the normal force. The normal force $N$ is equal to the weight of the block since the block is on a horizontal surface. Here, $N = 82\ N$ and the force required to start the motion $F_{s}=64\ N$. Then $\mu_{s}=\frac{F_{s}}{N}$.

Step2: Calculate coefficient of static friction

Substitute the values into the formula: $\mu_{s}=\frac{64\ N}{82\ N}\approx0.78$.

Step3: Recall kinetic - friction formula

The formula for kinetic friction is $F_{k}=\mu_{k}N$, where $F_{k}$ is the force of kinetic friction, $\mu_{k}$ is the coefficient of kinetic friction, and $N$ is the normal force. The force required to move the block at a constant speed is the force of kinetic friction. Here, $F_{k} = 40\ N$ and $N=82\ N$. Then $\mu_{k}=\frac{F_{k}}{N}$.

Step4: Calculate coefficient of kinetic friction

Substitute the values into the formula: $\mu_{k}=\frac{40\ N}{82\ N}\approx0.49$.

Answer:

E. 0.78 and 0.49