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8.

Normal force on B by the floor = F

Force exerted by block A on B: F1

F - m_B g - F1 = - m_B * 2 => F = F1 + m_B g - 2 m_B

F1 - m_A g = - m_A * 2 => F1 = g/2 - 1 = 4 Newtons

=======================

9. component of weight/acceleration due to gravity along the inclined slope

= g Sin 30 = g/2

Component of weight/acceleration along the groove OA = g/2 Sin 30 = g/4

s = u t + 1/2 a t² => t² = 2 s / a

t² = 2 * 5 / (g/4) = 4 => t = 2 sec.

=================

10. m1 = 100 kg m2 = 60 kg

a1 = acceleration of 100 kg upwards. Rope is also moving with the same acceleration.

T - 100 kg * g = 100 kg * a1 => T = 1000+ 100 a1 ---- (1)

Acceleration of the person = a

relative acceleration of the man wrt rope = a + a1 = 5g/4

=> a = 5g/4 - a1 = 12.5 - a1 upwards

T - 60 kg * g = 60 kg * a

T = 600 + 60 a = 600 + 60 (12.5 - a1) = 1350 - 60 a1 ----- (2)

Solving the two equations a1 = 350/160 m/sec/sec

T = 1000 + 100 * 35/16 = 1218 N

======================

11

F = buoyancy force upwards remains same

F - M g = M a => F = M (g+a)

F - (M-m) g = (M-m) a' => a' = [ M g + Ma - Mg + mg ] / (M-m)

a' = (M a + m g)/(M-m)

===============================

12. T - 5 g = 5 a

10 g - T = 10 a

a = g/3

=================

13. F - m g Sin 37 = m a along the slope

5.5 - 0.50 * g * 0.6 = 0.50 a => a = 5 m/sec/sec

v^2 = 2 a s = 2 * 5 * 10 = 100

v = 10 m/s at the top of the incline.

angle of projection = 37 deg.

y initial = 10 Sin 37 = 6 meters

Time to reach the top point : u Sin 37 / g = 10 * 0.6 / 10 = 0.60 sec

height reached = s = u sin 37 * 0.60 - 1/2 * 10 * 0.60^2

= 10 * 0.6 * 0.60 - 5 * 0.60^2 = 1.80 meters

total height at the top of the flight = 6 + 1.80 = 7.80 m

time to reach the ground = t => t^2 = 2s / g => t = 2 * 7.80 / 10 = 1.56 sec

total time of the flight = 2.16 sec.

total horizontal distance travelled after projection from the ramp

= u cos 37 * 1.56 sec

= 12.48 meters

Normal force on B by the floor = F

Force exerted by block A on B: F1

F - m_B g - F1 = - m_B * 2 => F = F1 + m_B g - 2 m_B

F1 - m_A g = - m_A * 2 => F1 = g/2 - 1 = 4 Newtons

=======================

9. component of weight/acceleration due to gravity along the inclined slope

= g Sin 30 = g/2

Component of weight/acceleration along the groove OA = g/2 Sin 30 = g/4

s = u t + 1/2 a t² => t² = 2 s / a

t² = 2 * 5 / (g/4) = 4 => t = 2 sec.

=================

10. m1 = 100 kg m2 = 60 kg

a1 = acceleration of 100 kg upwards. Rope is also moving with the same acceleration.

T - 100 kg * g = 100 kg * a1 => T = 1000+ 100 a1 ---- (1)

Acceleration of the person = a

relative acceleration of the man wrt rope = a + a1 = 5g/4

=> a = 5g/4 - a1 = 12.5 - a1 upwards

T - 60 kg * g = 60 kg * a

T = 600 + 60 a = 600 + 60 (12.5 - a1) = 1350 - 60 a1 ----- (2)

Solving the two equations a1 = 350/160 m/sec/sec

T = 1000 + 100 * 35/16 = 1218 N

======================

11

F = buoyancy force upwards remains same

F - M g = M a => F = M (g+a)

F - (M-m) g = (M-m) a' => a' = [ M g + Ma - Mg + mg ] / (M-m)

a' = (M a + m g)/(M-m)

===============================

12. T - 5 g = 5 a

10 g - T = 10 a

a = g/3

=================

13. F - m g Sin 37 = m a along the slope

5.5 - 0.50 * g * 0.6 = 0.50 a => a = 5 m/sec/sec

v^2 = 2 a s = 2 * 5 * 10 = 100

v = 10 m/s at the top of the incline.

angle of projection = 37 deg.

y initial = 10 Sin 37 = 6 meters

Time to reach the top point : u Sin 37 / g = 10 * 0.6 / 10 = 0.60 sec

height reached = s = u sin 37 * 0.60 - 1/2 * 10 * 0.60^2

= 10 * 0.6 * 0.60 - 5 * 0.60^2 = 1.80 meters

total height at the top of the flight = 6 + 1.80 = 7.80 m

time to reach the ground = t => t^2 = 2s / g => t = 2 * 7.80 / 10 = 1.56 sec

total time of the flight = 2.16 sec.

total horizontal distance travelled after projection from the ramp

= u cos 37 * 1.56 sec

= 12.48 meters