Answer:
d = 5.9 m
Explanation:
During the initial push, there are two forces acting on the box along the ramp: the component of gravity force along the ramp (directed to the bottom of the ramp), and the pushing force (up the ramp).
As we have as a given, the time during which the pushing force is applied, we can use Newton's 2nd Law, expressed in its original form, as follows:
Fnet *Δt = Δp = m* (vf-v₀) (1)
Fnet, in this case, is the difference between Fapp, and the projection of Fg along the ramp, which is equal to Fg times the sinus of the angle of the ramp with respect to the horizontal.
We choose as positive, the direction up the ramp, so we can write the follwing equation:
⇒ Fnet = Fapp - Fg*sin θ = 140 N - (7kg*9.8m/s²*sin 30º) = 140 N - 34.3 N
⇒ Fnet = 105.7 N
Replacing in (1):
105.7 N * 0.5 s = 7 kg* vf (v₀=0, as the box starts from rest)
Solving for vf:
vf = 105.7 N* 0.5 s / 7 kg = 7.6 m/s
When the push ends, the only force remaining along the ramp, is the component of Fg that we have already obtained, that will cause the box to have a deceleration, which we can find out aplying Newton's 2nd Law, as follows:
m*g*sin 30º = m*a
As we have defined as positive direction the one up the ramp, a will be negative (as it is slowing down the box) , and can be calculated as follows:
a = -34.3 N / 7 kg = -4.9 m/s²
As this value is constant, we can use any kinematic equation in order to get the distance traveled, farther the point where it disappeared the influence of the pushing force:
vf² - v₀² = 2*a*d
As we know that finally the box will come momentarily at rest (before falling under the influence of gravity) , we have vf =0:
⇒ -v₀² = 2*a*d
For this part, v₀, is just the value for vf, that we got above:
v₀= 7.6 m/s
⇒ -(7.6)² =2*(-4.9 m/s²)*d
Solving finally for d (the answer we are looking for):
d = (7.6)² (m/s)² / 2*4.9 m/s² = 5.9 m
