1) A 10.0 kg block on a table is connected by a string to a 63 kg mass, which is

Question

1) A 10.0 kg block on a table is connected by a string to a 63 kg mass, which is hanging over the edge of the table. Assuming that frictional forces may be neglected, what is the magnitude of acceleration of the 100 kg block when the other block is released? (See the figure.)

A) 8.5 m/s2 B) 9.0 m/s2 C) 7.5 m/s2 D) 8.1 m/s2

2) A 15 kg block is on a ramp that is inclined at 20

A 10.0 kg block on a table is connected by a string to a 63 kg mass, which is hanging over the edge of the table. Assuming that frictional forces may be neglected, what is the magnitude of acceleration of the 100 kg block when the other block is released? (See the figure.) 8.5 m/s2 B) 9.0 m/s2 C) 7.5 m/s2 D) 8.1 m/s2 A 15 kg block is on a ramp that is inclined at 20 degree above the horizontal. It is connected by a string to a 19 kg mass that hangs over the top edge of the ramp. Assuming that frictional forces may be neglected, what is the magnitude of the acceleration of the 19 kg block? (See the figure.) A) 4.5 m/s2 B) 3.8 m/s2 C) 4.2 m/s2 D) 4.0 m/s2

Explanation / Answer

1

Balancing forces for 63 kg :-

63g - T = 63a -----------(i)

Balancing forces for 10 kg:-

T = 10a -------------(ii)

By putting the value of T from (ii) in to (i) :-

=>63g - 10a = 63a

=>63g = 73a

=>a = 63 x 9.8/73 = 8.5 m/s^2

=>(d)

2

Net force=m2g-m1gsinA

=19*9.81-15*9.81*sin20=136.062 N

Acceleration=F/(m1+m2)=136.062/(15+19)=4.00 m/s^2

3

Ok there are two masses on one side of the pully B and C. B is on top and C is on bottom, the Force exerted on B by C is 60N. Mass of B is 18kg. Then there is a mass on the other side which has a mass of 12Kg.

Ok i tired this problem but i keep getting the wrong answer.

The total force on B is

T - (60 N + wB) = maa

and then the total force on B is

T - wA = -maa

Since A is accelerating downwards it has a negative acceleration and B has a positive accleration.

The answer should be 190N but i cant seem to get the answer. What am i doing wrong.

Thanks

4

Since Joe is pulling with a force of 200N and there is no acceleration (constant velocity), the net force is zero.

The kinetic frictional force should be equal to 200N

4

Let:

m be the mass,

k be the spring constant,

x be the extension,

a be the acceleration.

kx = ma

m = kx / a.

m = 45 * 0.88 / 2.1

= 18.9 kg to 3 sig. fig.

19kg is the closest answer of those listed

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