1. Two blocks (5.0 kg and 6.0 kg) are connected through a frictionless system. The angle of the ramp is 45°
A. Find the tension in the string.
B. Determine the normal force on the 5.0 kg block.

A) 45.5N b) 34.6N

Explanation:

Given the following :

Mass of block A(m1) = 5kg

Mass of block B (m2) = 6kg

Angle of incline (Θ) = 45°

Force pulling the down the incline = m1×a×sinΘ

Tension (T) = (m2×a) - (m2)×g

Where g = acceleration due to gravity

a = acceleration

Net force = (5×9.8×sin45°) - (6×9.8)

Net force = 34.648232 - 58.8

The net force acting on the body = 24.4N

Therefore,

Acceleration, a = Net force/ total mass

a = 24.4 / (6+5)

a = 2.22ms^-2

T = (m2 ×g) - (m2×a)

T = (6 × 9.8) - (6 × 2.22)

T = 58.8 - 13.32

T = 45.48

T = 45.5N

B) Normal reaction:

Horizontal component:

m1gCosΘ = 5 × 9.8 × cos45°

= 5 × 9.8 × 0.7071067

= 34.648232

= 34.6N

we need to know the time interval during which the force acts. impulse of force = momentum change

more than, less

less than, more