Answer
(a) W = 660 J
(b) W = -118 J
(c) The increase in potential energy is 353 J.
(d) The increase in kinetic energy is 189 J.
(e) The increase in kinetic energy is 189 J. This increase in the oven's kinetic energy is the same as we calculated in part (d).
Work Step by Step
(a) $W = F~d = (110~N)(6.00~m)$
$W = 660~J$
(b) $W = -F_f~d$
$W = -mg~cos(\theta)~\mu_k~d$
$W = -(10.0~kg)(9.80~m/s^2)~cos(36.9^{\circ})(0.250)(6.00~m)$
$W = -118~J$
(c) We can find the increase in potential energy.
$U_{grav} = mgh$
$U_{grav} = mgd~sin(\theta)$
$U_{grav} = (10.0~kg)(9.80~m/s^2)(6.00~m)~sin(36.9^{\circ})$
$U_{grav} = 353~J$
The increase in potential energy is 353 J.
(d) $K_2 + U_{grav,2} = K_1 + U_{grav,1}+\sum W$
$K_2 - K_1 = 660~J-118~J-353~J$
$K_2 - K_1 = 189~J$
The increase in kinetic energy is 189 J.
(e) $\sum F = ma$
$F - mg~sin(\theta) - mg~cos(\theta)~\mu_k = ma$
$a = \frac{F - mg~sin(\theta) - mg~cos(\theta)~\mu_k}{m}$
$a = \frac{110~N - (10.0~kg)(9.80~m/s^2)~sin(36.9^{\circ}) - (10.0~kg)(9.80~m/s^2)~cos(36.9^{\circ})(0.25)}{10.0~kg}$
$a = 3.157~m/s^2$
We can find the oven's speed.
$v^2 = v_0^2 + 2ax = 0 + 2ax$
$v = \sqrt{2ax} = \sqrt{(2)(3.157~m/s^2)(6.00~m)}$
$v = 6.155~m/s$
We can find the kinetic energy of the oven.
$K = \frac{1}{2}mv^2$
$K = \frac{1}{2}(10.0~kg)(6.155~m/s)^2$
$K = 189~J$
This increase in the oven's kinetic energy is the same as we calculated in part (d).
