A trunk of mass 15 kg is on the floor. The
trunk has a very small initial speed.
The acceleration of gravity is 9.8 m/s2 .
What constant horizontal force pushing the
trunk is required to give it a velocity of 10 m/s
in 20 s if the coefficient of sliding friction
between the trunk and the floor is 0.58?
Answer in units of N
EXPLAIN ALSO??
ANOTHER QUESTION
1300 N a push of the box to the right
236 kg of the box
μ = 0.44
A crate is pushed horizontally with a force.
The acceleration of gravity is 9.8 m/s2 .
Calculate the acceleration of the crate.
Answer in units of m/s2
1. Wt = m*g = 15kg * 9.8N/kg = 147 N. =
Weight of trunk.
Fk = u*mg = 0.58 * 147 = 85.26 N.=Force
of kinetic friction.
2. Fap = 1300 N.?
Mass = 236 kg
u = 0.44
Wc = m*g = 236kg * 9.8N/kg = 2313 N. =
Weight of crate.
Fk = u*mg = 0.44 * 2313 = 1018 N.
a=(Fap-Fk)/m=(1300-1018)/236=1.19 m/s^2
To calculate the constant horizontal force required to give the trunk a velocity of 10 m/s in 20 s, we need to consider both the force applied and the frictional force acting against the motion.
1. Determine the net force required:
Using Newton's second law, F_net = ma, where F_net is the net force, m is the mass of the trunk, and a is the acceleration of the trunk.
Given mass of the trunk (m) = 15 kg
Desired velocity (v) = 10 m/s
Time (t) = 20 s
Acceleration (a) = change in velocity / time = (v - 0) / t = 10 m/s / 20 s = 0.5 m/s²
Using F_net = ma, we can substitute the values:
F_net = 15 kg * 0.5 m/s² = 7.5 N
2. Determine the frictional force:
The frictional force (f_friction) is given by the coefficient of sliding friction (μ) multiplied by the normal force (f_normal).
The normal force (f_normal) is equal to the weight of the trunk, which is given by f_weight = mg, where g is the acceleration due to gravity.
Given the coefficient of sliding friction (μ) = 0.58
Acceleration due to gravity (g) = 9.8 m/s²
f_normal = mg = 15 kg * 9.8 m/s² = 147 N
f_friction = μ * f_normal = 0.58 * 147 N = 85.26 N
3. Calculate the required force (F_applied):
The required force is the sum of the net force and the frictional force.
F_applied = F_net + f_friction = 7.5 N + 85.26 N = 92.76 N
Therefore, the constant horizontal force required to give the trunk a velocity of 10 m/s in 20 s, considering the coefficient of sliding friction, is approximately 92.76 N.
For the second question regarding the acceleration of the crate:
We have:
Pushing force (F_applied) = 1300 N
Mass of the crate (m) = 236 kg
Coefficient of sliding friction (μ) = 0.44
Acceleration due to gravity (g) = 9.8 m/s²
1. Determine the frictional force:
The frictional force (f_friction) is given by the coefficient of sliding friction (μ) multiplied by the normal force (f_normal).
The normal force (f_normal) is equal to the weight of the crate, which is given by f_weight = mg, where g is the acceleration due to gravity.
f_normal = mg = 236 kg * 9.8 m/s² = 2312.8 N
f_friction = μ * f_normal = 0.44 * 2312.8 N = 1017.95 N
2. Determine the net force:
The net force (F_net) acting on the crate is equal to the sum of the pushing force (F_applied) and the frictional force (f_friction).
F_net = F_applied + f_friction = 1300 N + 1017.95 N = 2317.95 N
3. Calculate the acceleration (a):
Using Newton's second law, F_net = ma, we can solve for acceleration.
F_net = ma
2317.95 N = 236 kg * a
a = 2317.95 N / 236 kg ≈ 9.82 m/s²
Therefore, the acceleration of the crate is approximately 9.82 m/s².
To find the constant horizontal force required to give the trunk a velocity of 10 m/s in 20 seconds, we need to consider the forces acting on the trunk. There are two main forces to consider: the force pushing the trunk horizontally and the force of sliding friction between the trunk and the floor.
First, let's find the force of sliding friction. The equation for friction is given by:
friction = coefficient of friction * normal force
The normal force is the force exerted by the floor on the trunk in the vertical direction. Since the trunk is on the floor, the normal force is equal to the weight of the trunk, which is given by:
weight = mass * acceleration due to gravity
So, weight = 15 kg * 9.8 m/s² = 147 N
The force of sliding friction is then:
friction = 0.58 * 147 N = 85.26 N
Next, we can use Newton's second law of motion, F = m * a, to find the force required to accelerate the trunk to a velocity of 10 m/s in 20 seconds. Here, the acceleration of the trunk is the change in velocity divided by the time taken:
acceleration = (final velocity - initial velocity) / time
acceleration = (10 m/s - 0 m/s) / 20 s = 0.5 m/s²
Now, we can use Newton's second law to find the required force:
force = mass * acceleration
force = 15 kg * 0.5 m/s² = 7.5 N
However, this only considers the force required to overcome friction and accelerate the trunk. The total force required is the sum of the force of sliding friction and the force needed to accelerate the trunk:
total force = force of sliding friction + force to accelerate the trunk
total force = 85.26 N + 7.5 N = 92.76 N
Therefore, the constant horizontal force required to give the trunk a velocity of 10 m/s in 20 seconds, taking into account the coefficient of sliding friction, is approximately 92.76 N.
For the second question, to calculate the acceleration of the crate, we use the equation for the net force acting on an object:
net force = applied force - frictional force
Here, the applied force is given as 1300 N and the coefficient of friction is given as 0.44.
The frictional force is calculated using the same formula as before: friction = coefficient of friction * normal force.
Since the normal force is equal to the weight of the crate (mass * acceleration due to gravity):
normal force = 236 kg * 9.8 m/s² = 2312.8 N
Now we can calculate the frictional force:
frictional force = 0.44 * 2312.8 N = 1018.51 N
Using the equation for net force:
net force = applied force - frictional force
net force = 1300 N - 1018.51 N = 281.49 N
We can now use Newton's second law to find the acceleration:
acceleration = net force / mass
acceleration = 281.49 N / 236 kg = 1.19 m/s²
Therefore, the acceleration of the crate is approximately 1.19 m/s².