A block of mass 1 kg rests on a tabletop with coefficient of kinetic friction equal to 0.3. The block is?

Question

connected by a string which passes over a frictionless pulley to a second block of mass 2 kg which hangs vertically from the string. In the above problem suppose that a downward force of 5 N is applied to the 1 kg mass. The acceleration of the two blocks is

Answer 1

The system is subjected to negative friction Force Ff, and positive weight of 2 kg block Fw and 5 N

-Ff + mg + 5 = -0.3*1*g + 2g + 5 = 1.7g + 5 = 21.66 N

The system has mass 1+2 = 3kg

A = F/m = 21.66N/3 = 7.22 m/s^2

Answer 2

16.7 meters/sec squared. That is assuming the blocks are moving, not just sitting there. if they are not moving the block on the table is subject to static friction, not kinetic or dynamic friction and static friction is always higher than that for dynamic friction. It's a basic F=ma problem. Force is measured in Newtons, mass in kg and acceleration in meters per second squared. Acceleration due to gravity is 9.8 m/secsq. The hanging block can only exert a force of (2kg)(9.8M/secsq)=19.6N that is the tension in the string. The resistive force due to friction for the block on the tabletop is F=(normal force)(friction coefficient), so that is (9.8)(0.3)=2.94. The total force exerted on the tabletop block is therefore the tension exerted by the string minus the resistance of the dynamic friction, so 19.6-2.94=16.7Newtons. Acceleration is equal to force divided by mass, so the block on the table is 16.7 newtons divided by 1 kg=16.7 m/secsq. But again, that's assuming the blocks are actually moving. If not, all bets are off.

Answer 3

do a free body diagram of each mass

the forces working on the block on the table are
in the y direction : the weight of the block and the normal force the table is putting on the block, which are equal to each other

in the x direction: the tension of the rope and the friction force, which is the normal force multiplied by the friction coefficient

for this problem the normal force is the mass of the block multiplied by gravity plus the force applied to it(5 N)

so N=1kg*9.81m/s^2 + 5N or 9.81+5 = 14.81

so the friction force = 14.81*.3

on the second block, the forces are

in the y direction: the tension pulling up, and the weight of the mass pulling down


next step is to equate force to mass times acceleration, or F=ma

the acceleration of the blocks will be the same, only in different directions, there is no acceleration of the first block in the y direction, and there is no acceleration of the second block in the x direction.

for the first block sum the forces, then equate it to mass*acceleration, so this equation will look like

x: Sum of forces in the x direction= Tension - friction force= m(of block 1)*acceleration


the second block's equation is

y: Sum of forces in y direction = Tension - mass(block2)*gravity= m(block 2) * acceleration

this gives you 2 equations with 2 unknowns: tension and acceleration

you can solve this system to find both

Answer 4

Please convert 2.20 Cm to 0.022 m. Spring pressure = 88.2*0.022 N = a million.ninety 4 N. The spring pressure works alongside the horizontal axis = a million.ninety 4 cos 14.6 N = a million.88 N. typical pressure the block agains floor = 4.24*10 N = 40 2.4N. (assume gravity = 10 m/s²) Coefficient of kinetic friction = a million.88/40 2.4 = 0.40 4

Answer 5

Zero. It's resting on a table and won't move downward.