Answer:
F = force
f = friction
u = coefficient of friction
R = normal reaction force
a = Acceleration
m = mass of block
g = gravity
f = uR
F = Ma
Say the block is moving to the right.
The 146N force thus acts to the right, and the friction force to the left, since it resists movement.
The 146N force acts to the right, but the horizontal component of it is 146 cos 50 = 93.84: So this is the force to the right.
Since F = uR and we're trying to find u, we need both F and R. R is easy to get since it is just m x g. This is in fact already given as the weight 350N. So R = 350.
The block is moving at a constant speed, so the force to the right must = the force to the left.
F = ma, so 93.84 - f = (350/g) x 0
This means f must be 93.84 also.
so we have f = uR,
93.84 = u x 350
so u = 0.268 or
0.27 to 2dp.
Hope you understand this.
Explanation:
Mujhe nhi pata sorry. Lekin mai bhut logo ko jaanti hoon jo iss savaal ka jawaab de sakte hai.
A) change in ht after 180m = 180 * sin(4-deg.) = 12.56m
net work done by gravity on the cyclist = mass * gravity * height diff.
= 85 * 9.8 * 12.56
= 10470J
= 10.5kJ
B) Kinetic energy = 1/2 * mass * vel.^2 = work done by gravity = 10470J
vel.^2 = 10470 * 2 / 85 = 246.4
vel. = 15.7m/s
Answer:No, the slope won't be too steep
Explanation:
Force is an external agency that causes a body to change its position while friction is a force that causes a body to slide over another. This force is called the frictional force (Ff). The force that causes a body to move is the moving force (Fm).
The slope will be too steep if the frictional force is greater than the moving force since the frictional force tends to oppose the moving force.
According to the explanation, we need to get Ff and compare with the moving force along the plane.
If Ff>Fm it means the slope will be too steep but if Ff<Fm, the slope won't be too steep and as such the body can easily be moved along the plane.
Resolving forces acting along the plane we have FmSintheta + Ff = Fm (FmSintheta and Ff are added because they act in the same direction along the plane)
Fm=50N, theta=5°
Imputing this into the formula to get Ff;
50sin5°+Ff =50
Ff= 50-50sin5°
Ff= 50-4.35
Ff= 45.65N
Since Ff<Fm, this means the slope is not too steep and as such the 30kg load can be moved along the plane easily.