Question

An object with mass 110 kg moved in outer space. When it was at location < 13, -18, -2 > its speed was 19.5 m/s. A single constant force < 250, 390, -220 > N acted on the object while the object moved to location < 19, -23, -5 > m. What is the speed of the object at this final location

Answer 1

Answer:

v = 21.4m/s

Explanation:

Given r1 = < 13, -18, -2 > m,

F = < 250, 390, -220 > N, r2 = < 19, -23, -5 > m

Where r1 and r2 are position vectors in space and F is the constant force vector in space.

r2 - r1 = < 19–13, -23–(-18), -5–(-2) > m

Δr = < 6, -5, -3 > m

F = m×a

a = F/m

m = 110kg (scalar quantity)

a = < 250/110, 390/110, -220/110 >

a = < 2.27, 3.55, -2 > m/s²

Since the force is constant, the acceleration is also constant. Therefore the equations of constant acceleration motion apply here.

V² = u² +2aΔr

The magnitude of the acceleration is calculated as follows

a = √(2.27² + 3.55² +(-2)²)

a = √(21.7554)

a = 4.66m/s²

Magnitude of Δr,

Δr = √(6² + (-5)² +(-3)²) = √(70) = 8.37m

Having calculated the magnitude of the acceleration and displacement, we can now calculate the final velocity.

u = initial velocity = 19.5m/s

v² = 19.5² + 2×4.66×8.37

v² = 458.25

v = √(458.25)

v = 21.4m/s