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this is the answer of your question
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shashankavsthi:
I have answwr only and i.e 0.24 m and acceleration=6m/sec^2
when cos37=5/4 then answer is correct
your method is accurate but you taken Component wrong this leads to wrong answer...if you used taken component wrt53° then in your calculation you have to put
f=mgcos53°=>2×10×3/5
but how cos37 will be 5/4?? tell this
mg sintheta x= 1/2kx^2
answer edit nahi ho raha
pic deletw krke doosri lga do...lekin mgsintheta=1/2kx^2 se x^2=0.24 aayega
option nahi aa 4aha
nii nii sahi hai !!! welldone
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Mass of the block, m = 1 kg
Spring constant, k = 100 N m–1
Displacement in the block, x = 10 cm = 0.1 m
At equilibrium:
Normal reaction, R = mg cos 37°
Frictional force, f = μ R = mg Sin 370
Where, μ is the coefficient of friction
Net force acting on the block = mg sin 37° – f
= mgsin 37° – μmgcos 37°
= mg(sin 37° – μcos 37°)
At equilibrium, the work done by the block is equal to the potential energy of the spring, i.e.,
mg(sin 37° – μcos 37°)x = (1/2)kx2
1 × 9.8 (Sin 370 – μcos 37°) = (1/2) × 100 × (0.1)
0.602 – μ × 0.799 = 0.510
∴ μ = 0.092 / 0.799 = 0.115
Spring constant, k = 100 N m–1
Displacement in the block, x = 10 cm = 0.1 m
At equilibrium:
Normal reaction, R = mg cos 37°
Frictional force, f = μ R = mg Sin 370
Where, μ is the coefficient of friction
Net force acting on the block = mg sin 37° – f
= mgsin 37° – μmgcos 37°
= mg(sin 37° – μcos 37°)
At equilibrium, the work done by the block is equal to the potential energy of the spring, i.e.,
mg(sin 37° – μcos 37°)x = (1/2)kx2
1 × 9.8 (Sin 370 – μcos 37°) = (1/2) × 100 × (0.1)
0.602 – μ × 0.799 = 0.510
∴ μ = 0.092 / 0.799 = 0.115
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