High School

After the block of ice melts down to 196 kg, the coefficient of friction with the ground changes to 0.25. She wraps a chain around the ice and drags it with a force of 631 N to the right, angled up 25° from the ground.

How fast will it accelerate?

Answer :

Final answer:

The acceleration of the block of ice is approximately 0.77 m/s^2.

Explanation:

To find the acceleration of the block of ice, we can use Newton's second law of motion, which states that the acceleration of an object is equal to the net force applied to it divided by its mass.

The net force on the block is the force of friction, which can be calculated using the formula F_friction = µ * F_normal, where µ is the coefficient of friction and F_normal is the normal force.

Once we have the net force, we can use the formula F_net = m * a to solve for the acceleration, where F_net is the net force and m is the mass of the block.

In this case, we are given the coefficient of friction and the mass of the block, so we can calculate the acceleration.

Using the given values, we have: coefficient of friction (µ) = 0.25, mass (m) = 196 kg, and force (F) = 631 N.

First, we need to find the normal force. Since the block is on a horizontal surface, the normal force is equal to the force of gravity acting on the block, which can be calculated using the formula F_gravity = m * g, where g is the acceleration due to gravity.

Using the known values, we have: g ≈ 9.8 m/s2.

Next, we can calculate the force of friction using F_friction = µ * F_normal.

Finally, we can use F_net = m * a to solve for the acceleration, where F_net is the net force and m is the mass of the block.

Using the known values, we have: F_net = F - F_friction = 631 N - (0.25 * F_normal).

From there, we can calculate the acceleration using F_net = m * a. Rearranging the formula, we have: a = F_net / m.

Using the known values, we have: acceleration (a) = (631 N - (0.25 * F_normal)) / 196 kg.

Substituting F_normal = m * g into the formula, we have: acceleration (a) = (631 N - (0.25 * 196 kg * 9.8 m/s2)) / 196 kg.

Simplifying the equation, we have: acceleration (a) = (631 N - 480.4 N) / 196 kg.

Therefore, the acceleration of the block of ice is approximately 0.77 m/s2.

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