Answer :
a) The work done on the oven by the force F is 1341.7.
b) The work on the oven by the friction force is 329J.
c) The increase in potential energy for the oven is 991J.
d) The increase in oven's kinetic energy is 21.71J.
e) K.E is 87.23 j
a)
[tex]W = F \times x[/tex]
[tex]W = 120 N \times (14.0 cos(37))[/tex]
[tex]W = 120 N \times (14.0 cos(37))[/tex]
[tex]W = 1341.71[/tex]
b) [tex]F_f=\mu_k N[/tex]
[tex]F_f=0.25*12\times9.8[/tex]
[tex]= 29.4 N[/tex]
[tex]W_f= F_f\times x[/tex]
[tex]W_f= 29.0\times 14 cos(37)[/tex]
[tex]W_f= 328.72 J = 329 J[/tex]
c) increase in the internal energy
[tex]U_2 = mgh[/tex]
[tex]= 12\times9.81\times14sin(37)[/tex]
[tex]=[/tex] [tex]991 J[/tex]
d) the increase in oven's kinetic energy
[tex]U_1 + K_1 + W_other = U_2 + K_2[/tex]
[tex]1341.71 J - 329 J - 991 J = K_2[/tex]
[tex]K_2 = 21.71 J[/tex]
e)[tex]F - F_f = ma[/tex]
[tex](120N - 29.4N ) / 12.0kg = a[/tex]
[tex]a = 7.55[/tex]
[tex]vf^2 = v0^2 + 2ax[/tex]
[tex]vf^2 = 2(7.55m/s)(14.0m)[/tex]
[tex]V_f = 14.5396m/s[/tex]
[tex]K = 1/2(mv[/tex]
[tex]K = 1/2(12.0kg)(14.5396m/s)[/tex]
[tex]K = 87.238J[/tex]
Answer:
Answered
Explanation:
a) What is the work done on the oven by the force F?
W = F * x
W = 120 N * (14.0 cos(37))
<<<< (x component)
W = 1341.71
b) [tex]F_f=\mu_k N[/tex]
[tex]F_f=0.25\times12\times9.8[/tex]
= 29.4 N
[tex]W_f= F_f\times x[/tex]
[tex]W_f= 29.0\times 14 cos(37)[/tex]
W_f= 328.72 J = 329 J
c) increase in the internal energy
U_2 = mgh
= 12*9.81*14sin(37)
= 991 J
d) the increase in oven's kinetic energy
U_1 + K_1 + W_other = U_2 + K_2
0 + 0 + (W_F - W_f ) = U_2 + K_2
1341.71 J - 329 J - 991 J = K_2
K_2 = 21.71 J
e) F - F_f = ma
(120N - 29.4N ) / 12.0kg = a
a = 7.55m/s^2
vf^2 = v0^2 + 2ax
vf^2 = 2(7.55m/s)(14.0m)
V_f = 14.5396m/s
K = 1/2(mv^2)
K = 1/2(12.0kg)(14.5396m/s)
K = 87.238J