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All posts related to Force

Nass Tuitions
Shared from SAT Mock - 2 (Stage-1) on Oct 05, 2018 3:08 PM

A body of mass 5 kg moving on a horizontal surface with an initial velocity of 6 m/s comes to rest after 3 sec. If one wants to keep this body moving on the same surface with a velocity of 6 m/s, the force required is

 
8 N
10 N
30 N
15 N
Please type your answer before submitting.
Nass Tuitions

Very good

Aswath
Shared from Force - 1 on Jul 22, 2016 9:14 PM

A ball drops vertically onto a floor with momentum p and then bounces repeatedly. The coefficient of restitution is e. The total momentum imparted by the ball to the floor is

 
p (1 + e)
p
p
Please type your answer before submitting.
Aswath

Why.please explain

Rachna
Shared from Force - 1 on Apr 22, 2015 3:47 PM

A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab. The static coefficient of friction between the block and the slab is 0.60, while the kinetic coefficient is 0.40. The 10 kg block is acted upon by a horizontal force of 100 N. If g = 9.8 m/s2, the resulting acceleration of the slab will be

 
0.98 m/s2
1.47 m/s2
1.52 m/s2
6.1 m/s2
Please type your answer before submitting.
Rachna

I am not able to get this !

Prannv Dhawan
Shared from Mock NTSE - 2 (SAT - 2014) on Oct 14, 2014 9:43 PM

In the given arrangement, the pulleys are fixed and ideal, the strings are light, m1 > m2 and S is a spring balance which is itself mass less. The reading of S (in units of mass) is

 
m1 – m2
(m1 + m2)
Please type your answer before submitting.
Ashwani Garg

Reading of the spring balance = restoring force of the spring
rnSince M1 >M2, there will be an acceleration in the system (which consists of two masses M1, M2, Massless spring balance, massless strings, fixed pulleys).
rnM1g - M2g = (M1 + M2) a
rna= (M1 – M2) g/(M1 + M2)
rnRestoring force, R= T1 =T2
rnThus, R= T1= M1 (g-a) = M1 (g- (M1 – M2) g/(M1 + M2)) =M1g (M1 + M2 – (M1 – M2))/ (M1 + M2)
rn = M1g (2M2)/ (M1 + M2) = 2M1M2g/ (M1 + M2)
rnThe reading of Reading of the spring balance (in unit of mass) is thus = 2M1M2/ (M1 + M2)
rn

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Nipun
Shared from Science Test - 13 on Aug 30, 2014 9:12 PM

A mass of 2 kg falls from a height of 40 cm on a spring of force constant 625 N/m. The spring is compressed by

 
16 cm
0.4 cm
0.01 cm
0.04 cm
Please type your answer before submitting.
Sai vinay

Given mass=2kg height=0.4m force=625 v=(2gh)^(1/2) = 8^1/2 KE when the mass just touches the spring = 1/2 x m x v^2 = 8 This energy is utilised in compressing the spring. So, 1/2 x k x X^2 = 8

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