Starting from rest at t = 0, a 5.0-kg block is pulled across a horizontal surface by a constant horizontal force having a magnitude of 12 N. If the coefficient of friction between the block and the surface is 0.20, at what rate is the 12-N force doing work at t = 5.0 s?
A) 0.13 kW
B) 0.14 kW
C) 0.12 kW
D) 26 W
E) 12 W

Question

Quizplus · Accepted Answer

The rate at which work is done (power) by the force is given by $P = Fv$, where $F$ is the force applied and $v$ is the velocity of the block. To find the velocity at $t = 5.0$ s, we first need to calculate the net force acting on the block and then use it to find the acceleration. The net force ($F_{net}$) is the applied force minus the frictional force ($F_{friction} = \mu mg$), where $\mu$ is the coefficient of friction, $m$ is the mass, and $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$).Given:- $F = 12 \, \text{N}$- $\mu = 0.20$- $m = 5.0 \, \text{kg}$- $g = 9.8 \, \text{m/s}^2$The frictional force is:$$F_{friction} = \mu mg = 0.20 \times 5.0 \times 9.8 = 9.8 \, \text{N}$$The net force is:$$F_{net} = F - F_{friction} = 12 - 9.8 = 2.2 \, \text{N}$$The acceleration ($a$) of the block is given by Newton's second law, $F = ma$, so:$$a = \frac{F_{net}}{m} = \frac{2.2}{5.0} = 0.44 \, \text{m/s}^2$$Using the equation of motion $v = u + at$ (where $u$ is the initial velocity, $a$ is the acceleration, and $t$ is the time), and knowing the block starts from rest ($u = 0$):$$v = 0 + (0.44)(5.0) = 2.2 \, \text{m/s}$$Finally, the power ($P$) is:$$P = Fv = 12 \times 2.2 = 26.4 \, \text{W}$$Rounding to the nearest whole number gives $26 \, \text{W}$, which corresponds to choice D.

Starting from Rest at T = 0, a 5