When the system is experiencing a uniformly accelerated motion, there are a set of equations to work from. In this case, work is energy which consist solely of kinetic energy. That is, 1/2*m*v2. First, let's find the final velocity.
a = (vf - v0)/t
2.6 = (vf - 0)/4
vf = 10.4 m/s
Then W = 1/2*(2100 kg)*(10.4 m/s)2
W = 113568 J = 113.57 kJ
Answer:
There is much more friction on the rough surface than there is on the smooth surface.
Explanation:
Answer:
55.66 m
Explanation:
While falling by 50 m , initial velocity u = 0
final velocity = v , height h = 50 , acceleration g = 9.8
v² = u² + 2gh
= 0 + 2 x 9.8 x 50
v = 31.3 m /s
After that deceleration comes into effect
In this case final velocity v = 17 m/s
initial velocity u = 31.3 m/s
acceleration a = - 61 m/s²
distance traveled h = ?
v² = u² + 2gh
(17)² = (31.3)² - 2x 61xh
h = 690.69 / 2 x 61
= 5.66 m
Total height during which he was in air
= 50 + 5.66
= 55.66 m
Answer:
195.168 m
Explanation:
To find the magnitude of the vector you can use the Pythagorean Theorem since you have the height and base and the vector is really just the hypotenuse
Pythagorean Theorem:

Plug values in

Simplify

Add the two values

Take the square root of both sides

Fnet = (mass) (acceleration)
= 11 kg x 3.7m/s^2
= 41 N