If the 0.4 kg spring were used to launch the toy horizontally across a felt rug, with a coefficient of kinetic friction being 0.35, what would be the velocity of the toy after traveling 0.5 meter beyond its launch point?
The max height of the pop up is 0.214 m.
the initial horizontal speed of the pop up toy when it’s released would be 2.05 m/s.
The spring constant is 200 N/m
To solve this problem, we can use the principle of conservation of mechanical energy.
Given:
Mass of the toy (m) = 0.4 kg
Coefficient of kinetic friction (μ) = 0.35
Distance traveled beyond the launch point (d) = 0.5 m
Max height of the pop up (h) = 0.214 m
Initial horizontal speed (v) = 2.05 m/s
Spring constant (k) = 200 N/m
First, let's find the initial potential energy (PE) of the toy when it reaches the maximum height:
PE = mgh
PE = (0.4 kg)(9.8 m/s^2)(0.214 m)
PE = 0.84208 J
Next, let's find the work done by the friction force when the toy travels the distance (d) beyond the launch point:
Work done by friction (W) = force of friction (f) × distance (d)
W = μmgd
W = (0.35)(0.4 kg)(9.8 m/s^2)(0.5 m)
W = 0.686 J
Now, let's find the initial kinetic energy (KE) of the toy when it is launched horizontally:
KE = 0.5mv^2
KE = 0.5(0.4 kg)(2.05 m/s)^2
KE = 0.8405 J
Since the only external force acting on the toy is friction, the net mechanical energy (E) is conserved:
E = PE + KE - W
E = 0.84208 J + 0.8405 J - 0.686 J
E = 0.99558 J
Finally, let's find the final velocity (v') of the toy when it travels the distance (d) beyond the launch point:
E = 0.5mv'^2
0.99558 J = 0.5(0.4 kg)(v')^2
(v')^2 = (2 × 0.99558 J) / (0.4 kg)
(v')^2 = 4.9889 J / 0.4 kg
(v')^2 = 12.47225 m^2/s^2
v' = sqrt(12.47225 m^2/s^2)
v' = 3.53 m/s
Therefore, the velocity of the toy after traveling 0.5 m beyond its launch point is approximately 3.53 m/s.