7. PE=0.5×700n/m×0.9m^2
0.9^2=0.81m
0.5×700×0.81= 283.5J
8. 2000=0.5×(x)×1.5m^2
1.5^2= 0.25
0.25×0.5=0.125
2000=0.125 (x)
2000/0.125=x
x=16000 n/m
9. 4000=0.5 (375 n/m)×(x)^2
0.5×187.5 (x)
4000/187.5=21.3333333333
Answer:
The horizontally applied force = 2360 N
Explanation:
<em>Force:</em> Force can be defined as the product of mass and acceleration. the S.I unit of force is Newton (N)
Fh = Fr + ma......... Equation 1
Where Fh = horizontally applied force, Fr = friction force, m = mass of the crate, a = acceleration of the crate.
<em>Given: m = 400 kg, a = 1 m/s²</em>
Fr = 1/2 W, W = mg ⇒W = 400×9.8 = 3920 N
∴Fr = 1/2(3920), Fr = 1960 N
Substituting these values into equation 1
Fh = 1960 + 400×1
Fh = 1960 + 400
Fh = 2360 N
Therefore the horizontally applied force = 2360 N
The initial velocity of Ms. Stafford is

, while her acceleration is

This is a uniform accelerated motion, so we can calculate the total distance travelled by Ms. Stafford in a time of

using the law of motion for a uniform accelerated motion: