Answer:
303.29N and 1.44m/s^2
Explanation:
Make sure to label each vector with none, mg, fk, a, FN or T
Given
Mass m = 68.0 kg
Angle θ = 15.0°
g = 9.8m/s^2
Coefficient of static friction μs = 0.50
Coefficient of kinetic friction μk =0.35
Solution
Vertically
N = mg - Fsinθ
Horizontally
Fs = F cos θ
μsN = Fcos θ
μs( mg- Fsinθ) = Fcos θ
μsmg - μsFsinθ = Fcos θ
μsmg = Fcos θ + μsFsinθ
F = μsmg/ cos θ + μs sinθ
F = 0.5×68×9.8/cos 15×0.5×sin15
F = 332.2/0.9659+0.5×0.2588
F =332.2/1.0953
F = 303.29N
Fnet = F - Fk
ma = F - μkN
a = F - μk( mg - Fsinθ)
a = 303.29 - 0.35(68.0 * 9.8- 303.29*sin15)/68.0
303.29-0.35( 666.4 - 303.29*0.2588)/68.0
303.29-0.35(666.4-78.491)/68.0
303.29-0.35(587.90)/68.0
(303.29-205.45)/68.0
97.83/68.0
a = 1.438m/s^2
a = 1.44m/s^2
Answer:
height from where rock was thrown is 27.916 m
Explanation:
speed = 7.50 m/s
θ = 30°
g= 9.8 m/s²
horizontal distance = 18 m
time require for vertical displacement

t = 2.8 sec
now for calculation of height
s = ut + 0.5 a t²
-h = v sinθ× t + 0.5 ×(-9.8)× (2.8²)
-h = 7.5 sin30°× 2.8 - 0.5 ×(9.8)× (2.8²)
-h = -27.916 m
h= 27.916 m
height from where rock was thrown is 27.916 m
Well, in some systems the atoms melt or burn or change state like from liquid to gases when they reach a certain temperature. This could decrease the energy in the system.
Hope I helped