Question

on a frictionlkess 20° inelined plane. A fioece of 16 N acting parallel vo the incline and up 非-kg block slides on a frictionless 20s e iline is applied to the block. What is kgblxapplied to the block. What is the acceleration of the blok? 20ms down the incline down the incline b 5.3 m/s up the incline ) 2.0 m/s up the incline 3.9 m/ down the incline e. 3.9 m/s up the incline A 1S-kg mass has an acceleration of (4.0i-3.0j) mis. Oaly two forces act on the mast. If one ⓒ 7. A 1s-kg ma . is (2.0i-1.4j) N, what is the magnitude of the other force ? a. 4.1 N b. 6.1 N e 5.1 N d. 7.1 N e. 2.4 N mass. If one of the forces S. The first of two identical boxes of mass m is siting on level ground. The makes a 20° angle with the ground. The normal force of the level ground on the first box is Nc force of the ramp on the second box is Na. Which statemegs is correet? a. N-N- mg b. N-mg,e- mg sin 20 l second box is sitting on a ramp that first box is Ng the normal -9. The horizontal surface on which the objects slides Determine F on which the objects slide is frictioniess. If M-2.0kg, the tension in string 1 is 12N. a. 25 N 20 N c. 30N d. 35 N e. 40 N C 10. The horizontal surface on which the objects slide is frietionles. If M- 1.0 kg and the magnitude of the force of the small block on the large block is 5.2 N, determine F 4M 2M a. 6.0 N 9.0N F- ((1.3) m4 7.8 N 4.8 N e. 4.1 N Are my answers correct here ? I'm trying to figure out my mistakes , thank you
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Answer #1

#6)

Net fore is given by

F_{net} = F - mgsin20

F_{net} =16 - 3*9.8*sin20 = 5.945N

a = \frac{F_{net}}{m}

a = \frac{5.945}{3} = 1.98 m/s^2\:upwards

#7

As we know

F_{net} = ma

2\hat i - 1.4\hat j + F = 1.5*(4\hat i - 3\hat j)

F = 4\hat i - 3.1\hat j

magnitude is given by

F = 5.06 N

#8

N_L = mg

N_R = mgcos\theta

#9

as per the Newton's law

T = (M+2M)a

12 = 6a

a = 2 m/s^2

Now

F_{net} = (M+2M + 2M)a

F_{net} =5*2*2 = 20 N

#10

By Newton's Law

F= 4Ma

5.2= 4*a

a = 1.3 m/s^2

Again

F = (2M + 4M)a

F = 6*1.3 = 7.8 N

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