The lawnmower accelerates in the positive horizontal direction, so that the net horizontal force is, by Newton's second law,
(70 N) cos(-50°) = <em>m</em> (1.8 m/s²)
where <em>m</em> is the mass of the lawnmower. Solve for <em>m</em> :
<em>m</em> = ((70 N) cos(-50°)) / (1.8 m/s²)
<em>m</em> ≈ 25 kg
The lawnmower presumably doesn't get lifted off the ground, so that the net vertical force is 0. If <em>n</em> is the magnitude of the normal force, then by Newton's second law,
<em>n</em> - <em>m g</em> + (70 N) sin(-50°) = 0
<em>n</em> = <em>m g</em> + (70 N) sin(50°)
<em>n</em> = (25 kg) (9.8 m/s²) + (70 N) sin(50°)
<em>n</em> ≈ 300 N
Answer:
A rocket with more mass will speed up more slowly, just as in the horizontal example, but there is another effect. The force of gravity is now acting in the opposite direction to the thrust, so the resultant force pushing the rocket upwards is also less.
Explanation:
Answer:
Height is 11.25m
Explanation:
<u>Given the following data;</u>
Initial velocity, u = 0
Final velocity, v = 15m/s
Acceleration due to gravity, g = 10m/s²
To find the height, we would use the third equation of motion;
Where;
- V represents the final velocity measured in meter per seconds.
- U represents the initial velocity measured in meter per seconds.
- a represents acceleration measured in meters per seconds square.
- S represents the displacement (height) measured in meters.
<em>Making S the subject, we have;</em>
But a = g = 10m/s²
<em>Substituting into the equation, we have;</em>
S = 11.25m
<em>Therefore, the ball will reach a height of 11.25m before it begins to fall. </em>
