A boy shoves his stuffed toy zebra down a friction-less chute, starting at a height of 1.07 m above the bottom of the chute and with an initial speed of 1.87 m/s. The toy animal emerges horizontally from the bottom of the chute and continues sliding along a horizontal surface with coefficient of kinetic friction 0.277. How far from the bottom of the chute does the toy zebra come to rest? Take g = 9.81 m/s2.
Hint: How does the work performed on the toy by the force of friction relate to the requested distance? How do the initial height and speed relate to the frictional work?
Energy conservation tells you that
M*g*H + M*Vo^2/2 = (mu)*M*g*X
Mass M cancels out. Solve for slide distance X.
mu = 0.277
g = 9.81 m/s^2
Vo = 1.87 m/s
H = 1.07;
k = 0.277
g = 9.81 m/s^2
Vo = 1.87 m/s
we know, M*g*H + M*Vo^2/2 = k*M*g*X
So, x = (g*H + (Vo)^2/2) / (mu * g)
Thank you so much!!! this really helped!!!
Why did the stuffed toy zebra decide to go on this wild adventure? It must have been tired of just sitting on a shelf, gathering dust. I can understand its desire for some excitement, but let's see how this playful adventure ends.
To find out how far the toy zebra comes to rest, we need to consider the work performed on it by the force of friction. Work is calculated by the formula: work = force × distance × cos(angle). In this case, the force is the frictional force, and we can assume the angle is 180 degrees since the force is acting in the opposite direction of motion.
Since the chute is frictionless, no work is done on the toy zebra as it slides down. This means that all the initial potential energy is converted into kinetic energy. So we have the equation: mgh = 0.5mv^2, where m is the mass, g is the acceleration due to gravity, h is the initial height, and v is the initial speed.
Simplifying, we get: gh = 0.5v^2. Plugging in the given values, we have: (9.81 m/s^2) × (1.07 m) = 0.5 × (1.87 m/s)^2.
Solving this equation, we find the value of v^2 to be approximately 3.482 m^2/s^2.
Now, let's analyze the toy zebra's motion along the horizontal surface. The force of kinetic friction will act on it, opposing its motion. The work done by friction is given by: work = force × distance × cos(angle). In this case, the angle is 0 degrees since the force of friction acts in the same direction as the displacement.
The work done by friction will be equal to the initial kinetic energy of the toy zebra (which was converted from its initial potential energy) since it comes to rest. So we have: work = 0.5mv^2. Rearranging, we get: distance = (0.5mv^2) / (force × cos(angle)).
The force of friction is calculated by the formula: force = coefficient of friction × normal force, where the normal force is equal to the weight of the toy zebra. So we have: force = μ × mg.
Let's plug in the given values: force = (0.277) × (m) × (9.81 m/s^2).
Now, we can substitute the force into the distance equation and solve for the distance: distance = (0.5mv^2) / ((μ × mg) × cos(0)).
Simplifying further, we get: distance = (0.5v^2) / (μ × g). Plugging in the given values, we have: distance = (0.5 × 3.482 m^2/s^2) / (0.277 × 9.81 m/s^2).
Calculating this, we find the distance to be approximately 0.608 meters.
So, the adventurous stuffed toy zebra comes to rest approximately 0.608 meters from the bottom of the chute. I hope it enjoyed its thrilling journey, even if it needed a break at the end!
To find the distance from the bottom of the chute where the toy zebra comes to rest, we can use the concept of work and energy.
First, let's consider the toy zebra's motion down the friction-less chute. Since there is no friction, the only force acting on the toy zebra is gravity. We can use the principle of conservation of energy to analyze this part of the motion.
The initial height of the toy zebra is 1.07 m, and its initial speed is 1.87 m/s. The potential energy (PE) at the top of the chute is converted into kinetic energy (KE) at the bottom of the chute. The total mechanical energy remains constant, so we have:
PE_top = KE_bottom
mgh = (1/2)mv^2
where m is the mass of the toy zebra, g is the acceleration due to gravity, h is the initial height, and v is the velocity at the bottom of the chute.
Since the mass cancels out, we can solve for v:
v = sqrt(2gh)
Now we have the velocity of the toy zebra as it emerges from the bottom of the chute. On the horizontal surface, the toy zebra experiences friction opposing its motion. The work done by friction is equal to the force of friction multiplied by the distance traveled:
W_friction = f_friction * d
The force of friction can be calculated using the coefficient of kinetic friction (μ) and the normal force (N). The normal force is equal to the weight of the toy zebra, which can be calculated by multiplying the mass (m) by the acceleration due to gravity (g).
N = mg
The force of friction (f_friction) can be found by multiplying the coefficient of kinetic friction (μ) by the normal force (N).
f_friction = μ * N
Since work (W) is equal to the force applied along the direction of motion multiplied by the distance traveled, we can rearrange the equation to solve for distance (d):
d = W_friction / f_friction
Here, W_friction is the work done against friction, which can be calculated using the change in kinetic energy (ΔKE). Since the toy zebra comes to rest, the final kinetic energy is zero. Therefore, ΔKE = -KE_bottom.
ΔKE = - (1/2)mv^2
Substituting the equation for ΔKE and f_friction into the equation for distance:
d = (- (1/2)mv^2) / (μ * N)
Now we have the equation to calculate the distance (d) from the bottom of the chute where the toy zebra comes to rest.
Remember to substitute the given values of the coefficients, acceleration due to gravity, initial height, and initial velocity of the toy zebra into the equation to get the final answer.