A 50 Kg box rests on a horizontal surface. The coefficient of Static friction between the box and surface is 0.30. The coeffient of Kinetic friction is 0.20. What is the frictional force on the box if a 140 N horizontal push is applied to it. If a 175 N horizontal push is applied to it. Thanks
friction=50*9.8*.3 static=147
so if a 140 push is given, it does not move, as friction opposes it with a push back of 140N
If 175N is applied, it slides, with friction force 50*9.8*.20 N force
Thanks a bunch Bob Alex
A blog is given a brief push up a 20.0° incline to give it an initial speed of 12.0m.s
A. How far along the surface does the blog slide before coming to rest
B. How much time does it take the blog to travel to its starting position.THANS.
Let's assume that there is no frictional force acting on the blog in the horizontal direction, and the only force acting on it is the force of gravity which is directed downwards along the inclined plane.
The force of gravity can be resolved into two components: one component that is parallel to the incline, and one component that is perpendicular to the incline.
F_parallel = mg sin(20°)
F_perpendicular = mg cos(20°)
where m is the mass of the blog, g is the acceleration due to gravity, and the angle of inclination is 20°.
The initial velocity of the blog is given as 12.0 m/s, and the final velocity is 0 m/s because the blog comes to rest.
Using the kinematic equations of motion, we can find the distance travelled by the blog before it comes to rest.
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance travelled.
The acceleration of the blog along the incline can be found by using the force equation:
F_parallel = ma
so, a = F_parallel/m
substituting the values, we get:
a = g sin(20°)
Using the equation of motion:
0^2 = 12.0^2 + 2a*s
substituting the value of acceleration, we get:
s = -12.0^2/(2*g*sin(20°))
solving for s, we get:
s = 12.2 m
So, the blog slides 12.2 m along the surface before coming to rest.
To find the time taken by the blog to travel to its starting position, we can use the equation of motion:
s = ut + (1/2)at^2
where u is the initial velocity, a is the acceleration, and s is the distance travelled.
Since the blog comes to rest at the end point, the final velocity is 0 m/s.
The distance travelled by the blog is equal to twice the distance it travelled before coming to rest, so
s = 2(12.2) = 24.4 m
Substituting the values, we get:
24.4 = 12.0t + (1/2)g sin(20°)t^2
simplifying, we get:
(1/2)g sin(20°)t^2 + 12.0t - 24.4 = 0
solving for t using the quadratic formula, we get:
t = 2.86 s
Therefore, it takes the blog 2.86 s to travel to its starting position.
Well, it seems like this box is not getting a break from all these pushes! Let's calculate the frictional force for both scenarios.
When a 140 N push is applied, we need to determine if the box is in a static or kinetic state. To do that, we compare the applied force (140 N) with the maximum static friction force. The maximum static friction force can be found by multiplying the coefficient of static friction (0.30) with the normal force, which is equal to the weight of the box (50 kg * 9.8 m/s²).
For 140 N:
Maximum static friction force = 0.30 * (50 kg * 9.8 m/s²)
Frictional force = 140 N
Now, let's see what happens when a 175 N push is applied. This force exceeds the maximum static friction force and causes the box to start moving. Once the box starts moving, it transitions from static to kinetic friction.
For 175 N:
Kinetic friction force = coefficient of kinetic friction * normal force
= 0.20 * (50 kg * 9.8 m/s²)
Frictional force = kinetic friction force
= 0.20 * (50 kg * 9.8 m/s²)
I hope this helps! Just remember, when it comes to friction, things can get quite slippery – or in this case, sticky!
To calculate the frictional force on the box, we need to consider the maximum static friction and the kinetic friction.
1. When a 140 N horizontal push is applied:
To determine if the box will move or not, we compare the applied force (140 N) to the maximum static frictional force.
The maximum static frictional force can be calculated using the formula:
Maximum static friction force = coefficient of static friction * Normal force
The normal force acting on the box is equal to the weight of the box, which is given by:
Normal force = mass * gravity
Plugging in the values, we have:
Normal force = 50 kg * 9.8 m/s^2
= 490 N
Maximum static frictional force = 0.30 * 490 N
= 147 N
Since the applied force (140 N) is less than the maximum static frictional force (147 N), the box will not move. Therefore, the frictional force on the box is equal to the applied force, which is 140 N.
2. When a 175 N horizontal push is applied:
Again, we compare the applied force (175 N) to the maximum static frictional force to determine if the box will move.
Using the same formula as before:
Maximum static frictional force = coefficient of static friction * Normal force
The normal force is still 490 N, as the weight of the box remains constant.
Maximum static frictional force = 0.30 * 490 N
= 147 N
This time, the applied force (175 N) is greater than the maximum static frictional force (147 N), which means the box will start moving. Once the box is in motion, the frictional force acting on it will change from static friction to kinetic friction.
The kinetic frictional force can be calculated using the formula:
Kinetic friction force = coefficient of kinetic friction * Normal force
Plugging in the values, we have:
Kinetic friction force = 0.20 * 490 N
= 98 N
Therefore, when a 175 N horizontal push is applied, the frictional force on the box is equal to the kinetic friction force, which is 98 N.