a 50 kg crate is at rest on a level surface the coefficients of friction are 0.3 and 0.4,what is the normal force acting on the crate, a man begins pushing horizontally on the crate how much force must the man exert to start the crate moving, how much force must the man exert to keep the crate moving once it is already in motion, how fast would the crate be moving if he applied that force for 5 seconds, what distance would the crate have covered during that 5 second interval
weight = normal force = m g
= 50 * 9.81 Newtons
static force = .4 * 50 * 9.81
moving force = .3 * 50 * 9.81
assume use entire static force or it will not accelerate
.1*50*9.81 = m a = 50 a
so
a = .981 m/s^2
v = a t = .981 * 5
To find the normal force acting on the crate, we can use the equation:
Normal force = Weight of the crate
Weight of the crate = mass of the crate × acceleration due to gravity
Weight of the crate = 50 kg × 9.8 m/s² = 490 N
Therefore, the normal force acting on the crate is 490 N.
To find the force required to start the crate moving, we can use the equation:
Force to start moving = coefficient of static friction × normal force
Force to start moving = 0.3 × 490 N = 147 N
Therefore, the man must exert a force of 147 N to start the crate moving.
To find the force required to keep the crate moving once it's already in motion, we can use the equation:
Force to keep moving = coefficient of kinetic friction × normal force
Force to keep moving = 0.4 × 490 N = 196 N
Therefore, the man must exert a force of 196 N to keep the crate moving.
To find how fast the crate would be moving if the man applied a force for 5 seconds, we need to consider the concept of acceleration. Calculating the acceleration requires the additional information of the crate's mass, which is given as 50 kg.
Using Newton's second law:
Force = mass × acceleration
Acceleration = Force / mass
Acceleration = 147 N / 50 kg = 2.94 m/s²
During the 5 seconds, if the crate starts from rest, the final velocity can be found using:
Final velocity = Initial velocity + (acceleration × time)
Initial velocity is 0 m/s since the crate starts from rest.
Final velocity = 0 + (2.94 m/s² × 5 s) = 14.7 m/s
Therefore, if the man applies a force for 5 seconds, the crate's final velocity would be 14.7 m/s.
To find the distance covered by the crate during the 5-second interval, we can use the equation of motion:
Distance = Initial velocity × time + (1/2) × acceleration × time²
Using the given values:
Distance = 0 m/s × 5 s + (1/2) × 2.94 m/s² × (5 s)²
Distance = 0 + 36.75 m = 36.75 m
Therefore, the crate would have covered a distance of 36.75 meters during the 5-second interval.
To answer these questions, we first need to calculate the normal force acting on the crate.
The normal force (N) is the force exerted by a surface to support the weight of an object resting on it. In this case, with the crate at rest on a level surface, the normal force is equal in magnitude and opposite in direction to the weight of the crate.
Step 1: Calculate the weight of the crate. The weight (W) can be calculated using the formula: W = m * g, where m is the mass of the crate and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Given the mass of the crate (m) is 50 kg, we can calculate the weight: W = 50 kg * 9.8 m/s^2 = 490 N.
Step 2: The normal force (N) is equal to the weight (W). Therefore, the normal force acting on the crate is 490 N.
Now let's move on to the next part of the question.
To start the crate moving:
When an object is at rest, the force required to overcome static friction is the maximum static friction force (Fs). The formula to calculate the maximum static friction force is Fs = μs * N, where μs is the coefficient of static friction and N is the normal force.
Given the coefficient of static friction (μs) is 0.3, and the normal force (N) is 490 N, we can calculate the maximum static friction force: Fs = 0.3 * 490 N = 147 N.
Therefore, the man must exert a force equal to or greater than 147 N to start the crate moving.
To keep the crate moving once it is already in motion:
Once the crate is in motion, the force required to overcome the kinetic friction is the kinetic friction force (Fk). The formula to calculate the kinetic friction force is Fk = μk * N, where μk is the coefficient of kinetic friction and N is the normal force.
Given the coefficient of kinetic friction (μk) is 0.4, and the normal force (N) is 490 N, we can calculate the kinetic friction force: Fk = 0.4 * 490 N = 196 N.
Therefore, the man must exert a force equal to or greater than 196 N to keep the crate moving.
If the man applies that force for 5 seconds:
The force (F) exerted by the man will cause the crate to accelerate. The acceleration (a) can be calculated using Newton's second law: F = m * a, where F is the force, m is the mass of the crate, and a is the acceleration.
Given the mass of the crate (m) is 50 kg and the force exerted (F) is 196 N, we can rearrange the formula to solve for acceleration: a = F / m = 196 N / 50 kg = 3.92 m/s^2.
To find the final velocity (v) after 5 seconds, we can use the equation: v = u + a * t, where u is the initial velocity (which is 0 m/s since the crate starts from rest), a is the acceleration, and t is the time.
Plugging in the values, we have: v = 0 + 3.92 m/s^2 * 5 s = 19.6 m/s.
Therefore, the crate would be moving with a velocity of 19.6 m/s if the man applied that force for 5 seconds.
To find the distance covered during that 5-second interval:
We can use the equation: s = u * t + (1/2) * a * t^2, where s is the distance covered, u is the initial velocity, t is the time, and a is the acceleration.
Plugging in the values, we have: s = 0 * 5 s + (1/2) * 3.92 m/s^2 * (5 s)^2 = 0 + 49.0 m.
Therefore, the crate would have covered a distance of 49.0 meters during that 5-second interval.