The possible magnitude for the force of static friction on the stationary cart is 72.1 N.
The given parameters:
- <em>Applied force on the cart, F = 72.1 N</em>
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Based on Newton's second law of motion, the force applied to object is directly proportional to the product of mass and acceleration of the object.
F = ma
Static frictional force is the force resisting the motion of an object at rest.
where;
is the frictional force
Thus, the possible magnitude for the force of static friction on the stationary cart is 72.1 N.
Learn more about Newton's second law of motion: brainly.com/question/25307325
Answer:
Torque = 882Nm
Explanation:
Torque = Mg×distance
But plank's is pivoted ,therefore distance=3/2=1.5m
Mass of Nancy=60jg
Acceleration due to gravity, g=9.8m/s^2
Torque= 60×9.8×1.5
Torque= 882Nm
1. Gas
2. Hot
Are the answers.
Well because it has momentum and it needs to be slowed down.
Answer:
a) 0.64 b) 2.17m/s^2 c) 8.668joules
Explanation:
The block was on the ramp, the ramp was inclined at 20degree. A force of 5N was acting horizontal to the but not parallel to the ramp,
Frictional force = horizontal component of the weight of the block along the ramp + the applied force since the block was just about move
Frictional force = mgsin20o + 5N = 6.71+5N = 11.71
The force of normal = the vertical component of the weight of the block =mgcos20o = 18.44
Coefficient of static friction = 11.71/18.44= 0.64
Remember that g = acceleration due to gravity (9.81m/s^2) and m = mass (2kg)
b) coefficient of kinetic friction = frictional force/ normal force
Fr = 0.4* mgcos 20o = 7.375N
F due to motion = ma = total force - frictional force
Ma = 11.71 - 7.375 = 4.335
a= 4.335/2(mass of the block) = 2.17m/s^2
C) work done = net force *distance = 4.335*2= 8.67Joules
