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Physics question A spacecraft descends vertically near the surface of Planet X. An upward thrust of 25.0kN from its engines slows it down at a rate of 1.20 m/s^2, but if an upward thrust of only 10.0 kN is applied, it speeds up at rate of .80m/s^2.

Verified Answer :

syed514

F=ma and m is constant on any planet 
25000-mg=m1.2 
10000-mg=-m0.80 
mg is the weight 
25000/1.2-mg/1.2=m 
10000/0.80-mg/0.80=-25000/1.2+mg/1.2 solve for mg 
m*g=(10000/0.80+25000/1.2)/ (1/1.2+1/0.80) 
16 kN

The spacecraft's weight near the surface of Planet X is; 16 kN

What is the weight?

We want to find the spacecraft's weight near the surface of Planet X.

Formula for Force is;

F = ma

where;

m is mass and is constant on any planet

a is acceleration

Thus, by equilibrium;

T - mg = ma

we are given;

T = 25 kN = 25000 N

a = 1.2 m/s²

Thus;

25000 - mg = m 1.2   ---(eq 1)

Similarly, we have;

T = 25 kN = 25000 N

a = 1.2 m/s²

Thus;

10000 - mg = -m 0.80 ----(eq 2)

From eq 1, m = 25000/1.2 - mg/1.2    -----(eq 3)

From eq 2, -m = 10000/0.80 - mg/0.80   ----(4)

Put eq 3 in 4 to get;

(10000/0.80) - (mg/0.80) = -(25000/1.2) + (mg/1.2) solve for mg

multiply through by (1.2 0.8) to get;

12000 - 1.2mg = -20000 + 0.8mg

12000 + 20000 = (1.2 + 0.8)mg

32000/2 = mg

mg = 16000 N = 16 kN

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