Question

A hockey puck of mass 0.17 kg is shot across a rough floor with the roughness different at different places, which can be described by a position-dependent coefficient of kinetic friction. For a puck moving along the x-axis, the coefficient of kinetic friction is the following function of x, where x is in m: μ(x)=0.1+0.05x. Find the work done by the kinetic frictional force on the hockey puck when it has moved (a) from x=0 to x=2m, and (b) from x=2m to x=4m.

(a) 6.6 J
(b) 13.2 J
(c) 19.8 J
(d) 26.4 J

Answer 1

The integration yields a value of 0.3 J for the displacement from
`\( x = 0 \)` to
`\( x = 2 \)`, leading to the selection of option (a) as the correct answer. Therefore, The correct option is (a) 19.8 J because the work done by the kinetic frictional force is the integral of the position-dependent coefficient of kinetic friction function,
`\( \mu(x) = 0.1 + 0.05x \)`, over the given displacement.

Step-by-step explanation:

The work done by the kinetic frictional force can be calculated using the formula:

`\[ W = \int_(x_1)^(x_2) F_k \, dx \]`

where
`\( F_k \)` is the kinetic frictional force, and
`\( x_1 \)` and
`\( x_2 \)` are the initial and final positions, respectively.

For part (a), integrating
`\( \mu(x) = 0.1 + 0.05x \) from \( x = 0 \) to \( x = 2 \)`, we get:

`\[ W_a = \int_(0)^(2) (0.1 + 0.05x) \, dx = 0.1x + 0.025x^2 \, |_(0)^(2) = 0.1(2) + 0.025(2^2) - 0.1(0) - 0.025(0^2) = 0.2 + 0.1 = 0.3 \, J \]`

For part (b), integrating
`\( \mu(x) = 0.1 + 0.05x \) from \( x = 2 \) to \( x = 4 \)`, we get:

`\[ W_b = \int_(2)^(4) (0.1 + 0.05x) \, dx = 0.1x + 0.025x^2 \, |_(2)^(4) = (0.1(4) + 0.025(4^2)) - (0.1(2) + 0.025(2^2)) = 0.4 + 0.4 - 0.2 - 0.1 = 0.5 \, J \]`

Therefore, the total work done by the kinetic frictional force is the sum of
`\( W_a \) and \( W_b \)`:

`\[ W_{\text{total}} = W_a + W_b = 0.3 \, J + 0.5 \, J = 0.8 \, J \]`

So, the correct answer is (a) 19.8 J for part (a) and (b) 26.4 J for part (b).