Answer:
Explanation:
Given that,
height h= 1.83 m
Initial speed Vi = 1.91m/s
Coefficient of kinetic friction
uk = 0.269
According to the law of conservation of energy, states that sum of the potential energy and kinetic energy at top of the chute is equal to the kinetic energy at bottom.
∆P.E = ∆K.E
mgh= ½m(Vf²—Vi²)
m cancel out
gh = ½Vf² — ½Vi²
½Vf² = gh + ½Vi²
Vf² = 2gh + Vi²
Vf² = 2×9.81 × 1.83 + 1.91²
Vf² =39.5527
Vf = √39.5527
Vf = 6.29 m/s
The frictional force exerted on the toy from the bottom of the chute is
Fr = —uk•N
N = W =mg
Fr = ma
Then, ma = —uk•m•g
Divide through by m
a = -uk • g
a = -0.269 × 9.81
a = —2.64 m/s²
According to equation of motion to find the distance travel at the bottom
v² = u² + 2as
Note: the final velocity is zero, the body comes to rest. And the initial velocity is the velocity when the zebra reached the ground u =Vf = 6.29m/s
Then, v² = u² + 2as
0² = 6.29² + 2(-2.64)s
—6.29² = —5.278s
— cancels out
Then, s= 6.29²/5.27
s= 7.5m
The zebra moves 7.5m from the chute
Answer:
The zebra comes to rest at 4.58 m from the bottom of the chute
Explanation:
We are given;
Height; h= 1.83 m
Initial speed; Vi = 1.91 m/s
Coefficient of kinetic friction; = 0.269
Now, from the law of conservation of energy,
Initial Kinetic Energy + Initial Potential Energy = Work done by friction
Now,
Kinetic Energy is given by the formula, KE = (1/2)mv²
Potential Energy is given by the formula ; PE = mgh
Also, work done by friction is given as W = µmgd
Thus, we now have;
½mv² + mgh = µmgd
m will cancel out and we have;
½v² + gh = µgd
Plugging in the relevant values to get;
½(1.91)² + (9.8 x 1.83) = (0.44 x 9.8)d
1.82405 + 17.934 = 4.312d
d = 19.75805/4.312
d = 4.58 m
