HELP PLEEAAAASSSEEEEEEE What is the definition of net force?

Physics

1 answer:

aleksley [76]3 years ago

8 0

Answer:

the sum of all force being applied to an object.

Explanation:

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Assume your mass is 84 kg. The acceleration due to gravity is 9.8 m/s 2 . How much work against gravity do you do when you climb

KATRIN_1 [288]

M, mass=84 kg
height, h=3.9m
gravity, g= 9.8m/s2
W = F . d
F=force
d=Displacement
W=work done by force
Now by putting the values
F= m g (Acting downward )
d= h (Upward)
W= m g h ( work done against the force)
W= 84•9.8•3.9J
W= 3210.48
Therefore the answer will be 3210.48J.

7 0

2 years ago

A mug rests on an inclined surface, as shown in (Figure 1) , θ=17∘.

Anna [14]

Refer to the figure shown below.

g = 9.8 m/s², the acceleration due to gravity.
W = mg, the weight of the mug.
θ = 17°, the angle of the ramp.

Let μ = the coefficient of static friction.

The force acting down the ramp is
F = W sin θ = W sin(17°) = 0.2924W N
The normal reaction is
N = W cosθ = W cos(17°) = 0.9563W N
The resistive force due to friction is
R = μN = 0.9563μW N

For static equilibrium,
μN = F
0.9563μW =0.2924W
μ = 0.3058

The frictional force is F = μN = 0.2924W
The minimum value of μ required to prevent the mug from sliding satisfies
the condition
R > F
0.9563μW > 0.2924W
μ > 002924/.9563 = 0.306

Answer:
The frictional force is 0.2924mg, where m = the mass of the mug.
The minimum coefficient of static friction is 0.306

7 0

3 years ago

Read 2 more answers

As shown in the figure below, a bullet is fired at and passes through a piece of target paper suspended by a massless string. Th

NikAS [45]

Answer:

M = 0.730*m

V = 0.663*v

Explanation:

Data Given:

$v_{bullet, initial} = v\\v_{bullet, final} = 0.516*v\\v_{paper, initial} = 0\\v_{paper, final} = V\\mass_{bullet} = m\\mass_{paper} = M\\Loss Ek = 0.413 Ek$

Conservation of Momentum:

$P_{initial} = P_{final}\\m*v_{i} = m*0.516v_{i} + M*V\\0.484m*v_{i} = M*V .... Eq1$

Energy Balance:

$\frac{1}{2}*m*v^2_{i} = \frac{1}{2}*m*(0.516v_{i})^2 + \frac{1}{2}*M*V^2 + 0.413*\frac{1}{2}*m*v^2_{i}\\\\0.320744*m*v^2_{i} = M*V^2\\\\M = \frac{0.320744*m*v^2_{i} }{V^2} ....... Eq 2$