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
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
4 answers
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.
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.
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.
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