Physics: What is the potential energy of a spring that is compressed 0.65 m by a 25 kg block if the spring constant is 95 n/m? 1.6 j 7.9 j 15 j 20 j

(a) (b) (c) 1235.06N and 252.05 as a ratioExplanation:From Newton s second law of motionF=ma where m is mass, a is acceleration and F is net forceAlso from kinetic equation of motion, velocity and displacement are related using equationWhere v is final velocity, u is initial velocity, a is acceleration and s is displacementFrom the free body diagram attached, final velocity at maximum height is 0 and initial velocity is Also, the vertical component can be written as The negative sign before 2gs...

What is the potential energy of a spring that is

Ответ:

Hudman57

05.06.2020

Physics

20 J

Explanation:

The elastic potential energy stored in a spring is given by:

$U=\frac{1}{2}kx^2$

where

k is the spring constant

x is the compression/stretching of the spring

For the spring in the problem, we have:

k=95 N/m

x=0.65 m

Substituting into the equation, we find

$U=\frac{1}{2}(95 N/m)(0.65 m)^2=20 J$

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Ответ:

lilswetheart2007

28.10.2022

Physics

(a) $47.15ms^{-1}$

(b) $2470.13ms^{-2}$

Explanation:

From Newton's second law of motion

F=ma where m is mass, a is acceleration and F is net force

Also from kinetic equation of motion, velocity and displacement are related using equation

$v^{2}=u^{2}+2sa$

Where v is final velocity, u is initial velocity, a is acceleration and s is displacement

From the free body diagram attached, final velocity at maximum height is 0 and initial velocity is $usin80^{o}$

Also, the vertical component can be written as

$v_{y}^{2} }=u_{y}^{2} } -2gs$ The negative sign before 2gs means displacement is opposite the gravitational force

Where g is acceleration due to gravity, $u_{y}$ is vertical component of initial velocity and $v_{y}$ is vertical component of final velocity

Since $v_{y}$ is 0

$u_{y}^{2} } =2gs$

s=110 and g is taken as 9.8

$u_{y}=\sqrt{2*9.8*110}=46.43275$

$u_{y}=46.43ms^{-1}$

Also, it's evident that the vertical component of initial velocity is $u_{y}=u_{i}sin \theta$ where $\theta$ is angle of projection and $u_{i}$ is resultant velocity