A 10-kg dog is running with a speed of 5.0 m/s. what is the minimum work required to stop the dog in 2.40 s?
1 answer:
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Answer:
F = 9.81 [N]
Explanation:
To solve this problem we must use Newton's third le which tells us that the sum of forces on a body that remains static must be equal to one resulting from these forces in the opposite direction.
Let's perform a summation of forces on the vertical axis-y to determine the normal force N.
∑F = 0 (axis-y)
where:
m = mass = 4 [kg]
g = gravity acceleration = 9.81 [m/s²]
Now we know that the frictional force can be calculated using the following equation.
f = μ*N
where:
f = friction force [N]
μ = friction coefficient = 0.25
N = normal force = 39.24 [N]
Now replacing:
Then we perform a sum of forces on the X-axis equal to zero. This sum of forces allows us to determine the minimum force to be able to move the object in a horizontal direction.
∑F = 0 (axis-x)
If the coefficient was smaller, a smaller force (F) would be needed to start the movement, this can be easily seen by replacing the value of 0.25, by smaller values, such as 0.1 or 0.05.
If the coefficient were larger, a larger force would be needed.
Answer:
Minimum coefficient of kinetic friction between the surface and the block is .
Explanation:
Given:
Mass of the block = M
Spring constant = k
Distance pulled = x
According to the question:
<em>We have to find the minimum co-efficient of kinetic friction between the surface and the block that will prevent the block from returning to its equilibrium with non-zero speed. </em>
So,
From the FBD we can say that:
⇒ Normal force, <em>...equation(i)</em>
⇒ Elastic potential energy, =
<em> ...equation (ii)</em>
⇒ Frictional force, =
<em> ...equation (iii)</em>
⇒ Plugging (i) in (iii).
⇒
Now,
⇒ As we know that the energy lost due to friction is equivalent to PE .
⇒ <em>...considering PE as</em>
or
.
Arranging the equation.
⇒
⇒ <em>...eliminating x from both sides.</em>
⇒ <em>...dividing both sides wit Mg.</em>
Minimum coefficient of kinetic friction between the surface and the block is .
